Don't have a move yet, but just thinking out loud:
You know the 6 in the top right corner must be in the bottom row of that 3x3 square. You know this because the 6 in the upper left must come from the top two rows and the six from the upper middle square must be in the top two rows.
I feel like I'm explaining that really badly, but hopefully that makes sense.
that's the hard way to say it. the easy way is, "In the third row, the first available open space can't be a 6, as that column already has a 6, so the six must go in one of the other two available spaces"
EDIT: oh wait, i see, erase the other tiny 6's in the 3x3
EDIT: Edited pic, Took out the tiny other 6's in upper right 3x3
r5c3, r6c3, and r9c3 are the only cells in c3 that can be either 1, 3, or 7. Therefore, they cannot be anything other than 1, 3, or 7. That means that the only remaining cell in the bottom left that can be 5 is r8c3.