3x3::k:8960:8960:8960:6401:6401:6401:3074:1795:1795:8960:8960:6401:6401:2564:3074:3074:2309:2309:8960:6406:6151:6151:2564:3341:3341:3341:6425:6406:6406:6151:6151:6411:3338:3338:6425:6425:6406:2828:2828:6411:6411:6411:6425:3086:3086:6406:2575:3600:3600:6411:6411:6425:4625:4625:2575:2575:4370:3600:1299:1299:4884:4884:4625:2581:2070:4370:2583:3352:4872:4884:4884:4105:2581:2070:2583:2583:3352:4872:4872:4105:4105:

1. Outies n69 = r3c9 + r489c6 = +32(4)
-> Either r3c9 = 8 and r489c6 = {789}
or r3c9 = 9 and r489c6 = {689}
-> r4c7 is max 7

2. Innies n9 = r7c9,r9c7 = +10(2) (No 5)
r9c7 is min 2 -> r7c9 is max 8
Outies n6 = r3c9,r4c6,r7c9 = +23(6)
Since r7c9 cannot be 5 -> r7c9 is min 6.
I.e., r7c9 from (678), r9c7 from (432)

3! Whichever of (89) is in r3c9 goes in n6 in r56c8 -> goes in n9 in r78c7
Since 17(2)n7 = {89} also implies that value goes in n8 in r9c56
-> r7c9 cannot be 8 since that puts r9c7 = 2 and both (89) in n8 in r9 leaving no solution for 19(2)r8c6
-> r7c9 from (67)

4. Either: r3c9 = 9 puts remaining outies n6 = r4c6,r7c9 = +14(2)
This must be [86] since r7c9 from (67) and r4c6 from (689)

or: r3c9 = 8 puts remaining outies n6 = r4c6,r7c9 = +15(2) = [96] or [87]
But cannot be [96] since that puts both r4c7 and r9c7 = 4
-> must be r4c6,r7c9 = [87]

-> In both cases r4c6 = 8, -> r4c7 = 5

5. -> r89c6 from {69} or {79}
-> 13(2)n8 = {58}
-> 10(2)n2 not {28}

6. IOD n23 -> r3c49 = r2c3 + 14
-> r2c3 is max 3 and r3c4 is min 6

7. IOD n3 -> r3c9 = r23c6 + 4
-> Either r3c9 = 8 and r23c6 = {13}
Or r3c9 = 9 and r23c6 = {14} or {23}

8. Outies r12 = r3c15 = +8(2) (No 489)
-> 10(2)n2 cannot be [64]

9! Innies n1 = +10(3)
Either: 4 in r3 in r3c23 -> Innies n1 = [1{45}]
Or 4 in r3 in 13(3) -> 5 not in 13(3)
-> 5 in r3 only in r3c123
-> 5 in n3 only in r12c89
-> Either: 9(2)n3 = {45} and 7(2)n2 = {16}
Or 7(2)n2 = {25} and 9(2)n2 = {18}|{36}
-> 9(2)n3 cannot be {27}

10. 2 cannot be in r2c3 since that puts r3c49 = [79] and (HS 2 in n2) r3c6 = 2 which leaves no place for 4 in r3.
-> r2c3 from (13)

11! -> r23c6 cannot be {13} since that puts 10(2)n2 = [46] and 9(2)n3 = {45} (since {13} in r2c36 and it cannot be {27})
-> r3c9 cannot be 8
-> r3c9 = 9

12. -> r7c9 = 6 and r9c7 = 4
Also 9 in r56c8 and r78c7
-> r89c6 = [69]

13. Also 7 in n8 in c4
-> Remaining innies n5 = r46c4 = +12(2) = {39}
-> 18(3)r6c8 = [{48}6]
-> 12(2)n6 = [93]
-> 25(5)r3c9 = [9{1267}]

14. Also (HS 3 in n8) 5(2)n8 = {23} and r789c4 = {147}
-> r23c6 = {14}
-> 10(2)n2 = {37}
-> 5(2)n8 = [23]

15. Since r6c4 from (39) and 4 already in r6 -> 7 not in r7c4
-> r89c4 = {17} and r7c4 = 4
-> r9c3 = 2 and (IOD n7) r6c2 = 2
-> (HS 6 in n7)10(2)n7 = [46]
-> (HS 3 in n7) 8(2)n7 = {35}
-> r7c12 = {17}

16. Also (HS 5 in r7) r7c8 = 5
-> 8 in n9 in 16(3) (= {178})
-> 17(2)n7 = [89]
-> 19(4)n9 = [95{23}]
-> 8(2)n7 = [53]
-> 13(2)n8 = [85]

17. 2 in n1 in c1
-> Innies n1 must contain a 1
-> Outies r12 = r3c15 = [53]
-> Innies n1 = [3{16}]
-> (IOD n23) r3c4 = 8
Also -> r23c6 = [14]
-> (HS 1 in n3) 7(2)n3 = [61]
-> 9(2)n3 = [45]

18. Also (89) in n4 both in 25(5)r3c2
Since 2 not in 25(5) -> 25(5) = [1{3489}]
-> r3c3 = 6
-> (HS 6 in n5) 11(2)n4 = [65]
etc.

(By the way - really embarrassed I never noticed outies r89! Make some steps above much easier).

Prelims

a) R1C89 = {16/25/34}, no 7,8,9
b) R23C5 = {19/28/37/46}, no 5
c) R2C89 = {18/27/36/45}, no 9
d) R4C67 = {49/58/67}, no 1,2,3
e) R5C23 = {29/38/47/56}, no 1
f) R5C89 = {39/48/57}, no 1,2,6
g) R78C3 = {89}
h) R7C56 = {14/23}
i) R89C1 = {19/28/37/46}, no 5
j) R89C2 = {17/26/35}, no 4,8,9
k) R89C5 = {49/58/67}, no 1,2,3
l) 10(3) cage at R6C2 = {127/136/145/235}, no 8,9
m) 10(3) cage at R8C4 = {127/136/145/235}, no 8,9
n) 19(3) cage at R8C6 = {289/379/469/478/568}, no 1

1a. Naked pair in R78C3, locked for C3 and N7, clean-up: no 2,3 in R5C2, no 1,2 in R89C1
1b. 45 rule on N7 3 innies R7C12 + R9C3 = 10 = {127/145/235} (cannot be {136} which clashes with R89C1), no 6
1c. 45 rule on C12 2 outies R15C3 = 9 = {27/36/45}, no 1
1d. 45 rule on N1 3 innies R2C3 + R3C23 = 10 = {127/136/145/235}, no 8,9
1e. 10(3) cage at R6C2 = {127/145/235} (cannot be {136} = 6{13} which clashes with R7C56), no 6
1f. {127} cannot be 7{12} which clashes with R7C56), no 7 in R6C2
1g. 10(3) cage at R8C4 = {127/136/145/235}
1h. 7 of {127} must be in R89C4 (R89C4 cannot be {12} which clashes with R7C56), no 7 in R9C3
[Alternatively 45 rule on N7 1 innie R9C3 = 1 outie R6C2]
1i. 45 rule on R12 2 outies R3C15 = 8 = {17/26}/[53], no 4,8,9, no 3 in R3C1, clean-up: no 1,2,6 in R2C5
1j. 45 rule on N9 2 innies R7C9 + R9C7 = 10 = [19]/{28/37/46}, no 5, no 9 in R7C9
1k. 45 rule on N5 3 innies R4C46 + R6C4 = 20 = {389/479/569/578}, no 1,2
1l. 45 rule on N5 2 innies R46C4 = 1 outie R4C7 + 7 -> no 7 in R6C4 (IOU)
1m. 45 rule on R89 3 innies R8C378 = 14
1n. R8C3 = {89} -> R8C78 = 5,6 = {14/15/23/24}
1o. 45 rule on R89 3 outies R7C378 = 22 = {589/679}, 9 locked for R7

2a. 45 rule on N6 3(2+1) outies R37C9 + R4C6 = 23
2b. Max R37C9 = 17 -> min R4C6 = 6, clean-up: no 8,9 in R4C7
2c. Max R3C9 + R4C6 = 18 -> R7C9 = {678} (5 already eliminated), R9C7 = {234} (step 1i)
2d. R7C378 = 22 (step 1o), R7C9 = {678} -> R7C3789 = 28,29,30 = {5689/5789/6789}, 8 locked for R7
2e. 45 rule on N3 1 innie R3C9 = 2 outies R23C6 + 4
2f. Min R23C6 = 3 -> min R3C9 = 7
2g. Max R3C9 = 9 -> max R23C6 = 5 = {12/13/14/23}
2h. 45 rule on N23 2 innies R3C49 = 1 outie R2C3 + 14
2i. Max R3C49 = 17 -> max R2C3 = 3
2j. Min R3C49 = 15 -> R3C4 = {6789}
2k. 19(3) cage at R8C6 = {289/379/469/478} (cannot be {568} because R9C7 only contains 2,3,4), no 5
2l. R9C7 = {234} -> no 2,3,4 in R89C6

3a. 45 rule on N69 4(1+3) outies R3C9 + R489C6 = 32
3b. Max R489C6 = 24 -> min R3C9 = 8
3c. R3C9 = {89} -> R4C489 = 23,24 = {689/789}, 8,9 locked for C6

4a. 45 rule on N6 2 outies R37C9 = 1 innie R4C7 + 10 -> R4C7 less than either of R37C9
4b. R3C9 ‘sees’ all except R7C7 in C7 (R89C7 only contain 1,2,3,4), R3C9 = {89} -> R7C7 = {89}
4c. Naked pair {89} in R7C37, 8 locked for R7, R7C8 = 5 (step 1o), clean-up: no 2 in R1C9, no 4 in R2C9, no 7 in R5C9, no 2 in R9C7 (step 1j)
4d. R7C78 = [85/95] = 13,14 -> R8C78 = 5,6 = {14/23/24}
4e. Killer pair 3,4 in R8C78 and R9C7, locked for N9
4f. 19(3) cage at R8C6 (step 2k) = {379/469/478}
4g. Hidden killer pair 8,9 in R4C6 and R89C6 for C6, R89C6 contains one of 8,9 -> R4C6 = {89}, R4C7 = {45}
4h. 19(3) cage = {379/469} (cannot be {478} = {78}4 which clashes with R4C67 = [85/94]), 9 locked for C6 and N8 -> R4C67 = [85], clean-up: no 7 in R5C8
4i. R5C23 = {47/56}/[92] (cannot be [83] which clashes with R5C89), no 3,8, clean-up: no 6 in R1C3 (step 1c)
4j. R89C5 = {58} (cannot be {67} which clashes with R89C6), locked for C5, 5 locked for N8, clean-up: no 2 in R3C5, no 6 in R3C1 (step 1i)
4k. R4C6 = 8 -> R46C4 = 12 (step 1k) = [39/75/93]
4l. 18(3) cage at R6C8 = {279/369/468} (cannot be {189} because R7C9 only contains 6,7, cannot be {378} which clashes with R5C89), no 1
4m. R7C9 = {67} -> no 6,7 in R6C89
4n. 1,6,7 in N6 only in 25(5) cage at R3C9 = {12679/13678}, no 4
4o. R3C9 = {89} -> no 8,9 in R4C89 + R56C7

[An unexpected breakthrough step.]
5a. 45 rule on N8 2 outies R9C37 = 1 innie R7C4 + 2
5b. 45 rule on N7 1 outie R6C2 = 1 innie R9C3
5c. Consider whether or not R6C2 and R9C3 contain 5
R6C2 = 5 => R6C34 cannot total 8 => no 6 in R7C4
or R6C2 and R9C3 don’t contain 5 => max R9C37 = 7 => max R7C4 = 5
-> no 6 in R7C4
5d. R7C9 = 6 (hidden single in R7), R9C7 = 4 (step 1j), 16(3) cage at R8C9 = {178} (only remaining combination), 1,8 locked for N9, clean-up: no 1 in R1C8, no 3 in R2C8, no 4 in R6C2, no 6 in R8C1
5e. R7C3 = 9, 8 in C7 only in R123C7, locked for N3 -> R3C9 = 9 (or step 4b in reverse), clean-up: no 1 in R2C89, no 3 in R5C8
5f. R7C9 = 6 -> R6C89 = 12 = [48/84/93]
5g. Naked quad {3489} in R56C89, 3 locked for C9 and N6, clean-up: no 4 in R1C8, no 6 in R2C8
5h. R1C89 = [34/61] (cannot be [25] which clashes with R2C89)
5i. R2C9 = 5 (hidden single in C9) -> R2C8 = 4, R1C89 = [61], clean-up: no 6 in R3C5, no 2 in R3C1 (step 1i)
5j. R9C7 = 4 -> R89C6 = 15 = {69}, 6 locked for C6 and N8
5k. 7 in N8 only in R789C4, locked for C4, clean-up: no 5 in R6C4 (step 4k)
5l. Naked pair {39} in R46C4, locked for C4 and N5

6a. 3 in N8 only in R7C56 = {23}, locked for R7, 2 locked for N8
6b. R3C9 = R23C6 + 4 (step 2e)
6c. R3C9 = 9 -> R23C6 = 5 = [14] (cannot be {23} which clashes with R7C6), clean-up: no 9 in R2C5, no 7 in R3C1 (step 1i)
6d. Naked pair {37} in R23C5, locked for C5 and N2 -> R7C56 = [23], R1C5 = 9
6e. R2C6 = 1 -> R12C7 = 11 = {38}, locked for N3, 3 locked for C7 -> R8C78 = [23], R3C78 = [72], R23C5 = [73], R3C1 = 5 (step 1i), clean-up: no 4 in R5C3 (step 1c), no 7 in R5C2, no 7 in R9C1, no 5,6 in R9C2
6f. Naked pair {16} in R3C23, 6 locked for R3 and N1 -> R3C4 = 8
6g. Naked pair {25} in R1C46, 2 locked for R1, N2 and 25(5) cage at R1C4 -> R2C34 = [36], clean-up: no 6,7 in R5C3, no 4,5 in R5C2

7a. Naked pair {78} in R89C9, locked for C9 and N9
7b. R9C8 = 1 -> R9C4 = 7, R8C4 + R9C3 = 3 = [12], R5C3 = 5 -> R5C2 = 6, R1C3 = 4 (step 1c), R6C2 = 2 (step 5b)
7c. R3C23 = [16], R4C8 = 7 -> R4C3 = 1, R4C4 = 9 (cage sum)
7d. R5C7 = 1, R5C5 = 4, R5C9 = 3 -> R5C8 = 9

and the rest is naked singles.

Preliminaries from SudokuSolver
Cage 17(2) n7 - cells ={89}
Cage 5(2) n8 - cells only uses 1234
Cage 7(2) n3 - cells do not use 789
Cage 8(2) n7 - cells do not use 489
Cage 12(2) n6 - cells do not use 126
Cage 13(2) n8 - cells do not use 123
Cage 13(2) n56 - cells do not use 123
Cage 9(2) n3 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 10(2) n2 - cells do not use 5
Cage 11(2) n4 - cells do not use 1
Cage 10(3) n78 - cells do not use 89
Cage 10(3) n47 - cells do not use 89
Cage 19(3) n89 - cells do not use 1

This is a highly optimised solution so any clean-up needed is stated.
1. "45" on n69: 4 outies r3c9 + r489c6 = 32
1a. max. r489c6 = 24 -> min r3c9 = 8
1b. -> r489c6 = 23/24 = {689/789}(no 1...5): 8 & 9 locked for c6
1c. -> no 8,9 in r4c7
1d. min. r89c6 = {68} = 14 -> max. r9c7 = 5

2. "45" on r89: 3 innies r8c378 = 14
2a. r8c3 = (89) -> r8c78 = 6/5 (no 6,7,8,9)

3. r3c9 sees all 8,9 in c7 apart from r7c7 so must repeat there
3a. -> r7c7 = (89)

4. "45" on r89: 3 outies r7c378 = 22 = {89}[5]
4a. 8,9 both locked for r7

5. "45" on n9: 2 innies r7c9 + r9c7 = 10 = [73/64]

6. "45" on n69: 1 outie r3c9 = 2 innies r47c7
6a. -> r47c7 = 8/9 = [53/54/63]
6b. -> r4c6 = (78)

7. 19(3)r8c6 must have 9 for c6
7a. = {79}[3]/{69}[4](no 8)
7b. 9 locked for n8

8. hsingle 8 in c6 -> r4c6 = 8, r4c7 = 5
8a. no 7 in 12(2)n6

9. 18(3)r6c8 must have 6,7 for r7c9 = {279/369/378/468}(no 1)
9a. can't have both 6,7 -> no 6,7 in r6c89

10. 1,6,7 in n6 only in 25(5) = 167{29/38}(no 4)
10a. can't have both 8,9 -> no 8,9 in that cage in n6

11. hidden killer pair 8,9 in c7 -> r123c7 must have one of 8,9
11a. -> killer pair with r3c9: both locked for n3

12. 5 in n3 only in 7(2) = [25] or in 9(2) = [45]
12a. -> {34} blocked from 7(2) (Locking out cages)
12b. = {16}/[25](no 3,4; no 2 in r1c9)
12c. and {27} blocked from 9(2)
12d. = {36}/[45](no 1,2,7; no 4 in r2c9)
12e. and 6 locked in those two cages for n3

13. "45" on n23: 1 outie r2c3 + 14 = 2 innies r3c49
13a. min. two innies = 15 (no 1,2,3,4,5)
13b. max. two innies = 17 -> max. r2c3 = 3

14. 8 in n8 only in 13(2) = {58}: both locked for c5, 5 for n8
14a. no 2 in 10(2)n2

15. 17(2)n7 = {89}: both locked for c3 and n7
15a. no 1,2 in 10(2)n7

Andrew's really neat breakthrough combines my step 16 & 18. Missed that.
16. "45" on n8: 1 innie r7c4 + 2 = 2 outies r9c37 with r9c7 from (34)
16a. no 5 in r7c4 -> r9c37 <> 7 -> no 3,4 in r9c3
16b. max. r7c4 = 7 -> max. r9c37 = 9 -> no 7 r9c3
16c. min. r9c37 = 4 -> min. r7c4 = 2

17. 10(3)r8c4: {127/136/145/235}
17a. <163> blocked by 5(2)n8 needing 1 or 3
17b. -> no 6 in r9c3

18. "45" on n7: 1 outie r6c2 = 1 innie r9c3 = (125)
18a. note: -> r123c1 must also have (125)
18b. 5 in n7 only in 8(2) or r9c3 -> 5 locked for c2 in r6c2 or 8(2)

19. 11(2)n4: {38} blocked by 12(2)n6 which needs one of them
19a. = [92]/{47}/[65](no 3,8; no 2 in r5c2, no 6 in r5c3)

20. "45" on c12: 2 outies r15c3 = 9 = {27/45}(no 1,3,6)

Key steps
21. "45" on n1: 3 innies r2c3 + r3c23 = 10
21a. remembering that one of r123c1 + r6c2 + r9c3 = (125)
21b. -> {127} blocked from h10(3)n1 by 5 in r9c3 and {45} in r15c3
21c. and {145} blocked from h10(3)n1 by 2 in r9c3 and {27} in r15c3
21d. = {136/235}(no 4,7,8,9)
21e. 3 locked for n1

22. "45" on c12: 1 outie r1c3 + 2 = 1 innie r5c2
22a. and hidden pair 4,7 in r1456c3
22b. -> {47} blocked from 11(2)n5 since r1c3 = (25) -> both 4,7 would have to be in n4 for c3 which is not possible (Combo Crossover Clash CCC)
22c. -> 11(2)n4 = [65/92]
22d. -> r15c3 = [72/45]

23. "45" on r12: 2 outies r3c15 = 8 (no 4,8,9)

24. 4 in r3 only in 13(3) = {148/247/346}(no 5,9)

25. 5 in r3 only in h8(2)r3c15 = [53] or in h10(3)n1 -> 3 in r3c23 in h10(3) must also have 5 (Locking out cages) but this makes h10(3) = [235] which clashes with r5c3 = (25)
25a. -> no 3 in r3c23
25b. -> r2c3 = 3
25c. -> 9(2)n3 = [45]
25d. 7(2)n3 = {16}: 1 locked for n3: both locked for r1

26. 13(3)r3c6 = [4]{27}(only valid combination): 2 & 7 locked for r3 and n3

27. r3c5 = 3 (hsingle r3) -> r3c1 = 5 (h8(2))

28. r1c7 = 3 (hsingle n3)
28a. -> r2c67 = 9 = [18]

Much easier now.