Firstly, let's apply all "basic level" techniques - naked/hidden singles, naked/hidden subsets, box-line intersections (aka locked candidates):
Code:
+----------------+----------------+----------------+
| 6 5 1 | 4 3 7 | 8 9 2 |
| 3 8 2 | 69 69 5 |-147 147 17 |
| 9 7 4 | 2 8 1 | 56 56 3 |
+----------------+----------------+----------------+
| 15 46 59 | 3 579 8 | 2 46 #157 |
| 2 34 358 | 1 @57 6 |*47 38 9 |
| 158 369 7 | 59 2 4 |*16 38 #15 |
+----------------+----------------+----------------+
| 4 39 39 | 8 @15 2 |*157 157 6 |
| 7 1 56 | 56 4 3 | 9 2 8 |
| 58 2 568 | 7 156 9 | 3 15 4 |
+----------------+----------------+----------------+
r57c5 (marked @) prevents r57c7 to be [71].
r46c9 (marked #) prevents r56c7 to be [71].
As a result, r25c7 can't be [47] (since it will lead to r256c7 or r257c7 to be [471]).
Therefore r2c7 can't be 4. And the rest is solvable by singles.
This path is different to the "textbook solving path" as it uses only
one advanced move instead of two to crack the puzzle. However pedantic people will probably refer it as not elegant enough (i.e. feels too much like forcing chain).
Another unorthodox move I discovered is like this (back to the pencilmark grid as above):
Code:
+----------------+----------------+----------------+
| 6 5 1 | 4 3 7 | 8 9 2 |
| 3 8 2 | 69 69 5 | 147 147 17 |
| 9 7 4 | 2 8 1 | 56 #56 3 |
+----------------+----------------+----------------+
| 15 46 59 | 3 579 8 | 2 @46 157 |
| 2 34 358 | 1 #57 6 |@47 38 9 |
| 158 369 7 | 59 2 4 | 16 38 15 |
+----------------+----------------+----------------+
| 4 39 39 | 8 *15 2 | 157 157 6 |
| 7 1 56 | 56 4 3 | 9 2 8 |
| 58 2 568 | 7 156 9 | 3 *15 4 |
+----------------+----------------+----------------+
Naked triples {157} @ both r7c578 & r7c78+r9c8
=> r7c5+r9c8 (marked *) must be [11|55] (i.e. the 2 cells must be identical)
(This could be viewed as a mini version of Law of Leftovers, I call it "cell cloning".)
Now r4c8+r5c7 (marked @) can't be [44] (both @ n6)
=> r3c8+r5c5 (marked #) can't be [67]
=> r7c5+r9c8 (marked *) can't be [55], must be [11].
After that you still need
Børge's 2nd hint to crack the puzzle. But it is an alternative as his 1st hint (albeit probably not as simple/elegant).