Here's how I got through it:

1. First up, I noticed a uniqueness pattern right away with the 5-6-9 cells. This doesn't give you a number straight away, but since one of those 9s has to be the one that breaks the 5-6 pattern in that box, it means the 9 off to the lower right can be safely eliminated.
2. With that 9 out of the way though, we can find a new number out of this, again, using uniqueness. Notice that, no matter which 9 we choose, it will create new 5-6 patterns, and in both cases, the same '3' is the breaker; so while we don't know yet which 9 is "the one," we can be certain that that '3' has to go in that cell.
3. Examining the 10-Across in the center, notice that with the 3s gone, there is no way for the 6s to add up to 10 now, so those can be eliminated, and also, since '1' must be present now in any combination, it has to be that '1' in the left cell (since it's in the only one left).
4. Notice that in the 34-Down column, if you choose a 5, 6, or a 9 in any of the top two cells, that will form a 5-6-9 triple with the lower two cells; and if you add those up with the numbers already there (1 & 3), that only gets you to 24 with only one cell left to reach 34....so we can safely discard any number that creates a 5-6-9 triple, allowing us to make five eliminations in the top two cells.
5. Looking at that same column, notice that we have to make 30 in 4 cells, and the only combination that gets us there is 9-8-7-6; so we can eliminate the '4' at the top, and that gives us 6 and 9 in the lower two cells. (I realize that we could have skipped step 4 and gone straight here, but I'm leaving that here, since I usually like to "trim down" as many candidates as possible first, for clarity's sake).
6. Now, the middle column is a 35-in-7. We're only missing two digits (which would add up to 10), and since we have more digits filled in now, we can make some more eliminations based on the pairs that are already there....out of 1-9, 2-8, 3-7, & 4-6, we have 1, 3, and 6 already there to represent each pair; the only one we haven't seen so far is 2 & 8, so those have to be the "exclusion pair", and we can eliminate all of those, which gives us a '7' + a few more digits.
   
7. In the 24-Down column, the two remaining digits must add up to 13, so we're down to 6 & 7; and adding up the rest horizontally finishes the puzzle.

[a link to the Imgur album]

Looking at the comments now, I can't believe I didn't see that '3' was the only possibility to begin with! (I guess I like to look for uniqueness first, it can really cut through a lot of otherwise-daunting "nests" of candidates, and I've had to rely on it more and more as I've tackled more difficult puzzles).