Our focus concerns the contents of F6, but our considerations will lead us to learn what goes in F5, after which the rest of the puzzle is not hard.
A little thought shows that A8-A9 must contain 1 and 6, so A5-A6 must contain 2 and 5. A slightly more involved sequence--(The 5s in D1 and E9 imply that 5 must go somewhere in F4-F6. 5 is forbidden in G5, so it must go in H4-I4, thus eliminating F4)--shows that either F5 or F6 must contain a 5. Now comes the key point: F5-F6 cannot have a 2, since if it did both A5-A6 and F5-F6 would hold 2 and 5, and there would be duplicate solutions to the puzzle. Therefore F6 cannot be a 2.
Now let's try to find what does go in F6. Looking at row 6, remembering F6 cannot be 2, leads to F6 being 5, 7 or 9. Now 9 can only go in F6 or F8 in the F column. So let's try 9 in F8, hoping to eliminate either 5 or 7 from consideration in F6. A 9 in F8 requires that 7 goes in D8. But in row 6 the 7 can only go in D6 or F6. The only remaining possibility is F6. Thus we have learned that F6 can only be 9 or 7, not 5. But since either F5 or F6 must be 5, the only possibility is that F5 is a 5. By studying F6, we have the collateral result that F5=5.
Knowing F5=5, the puzzle solves easily. (One possible sequence of deductions is A6=5, A5=2, G5=4, F4=3...)
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