OK I'm not saying it doesn't work because I'm willing to be proved wrong, but I think there's a mistake.

Top left box...

Only way to make 4 with two digits is 1/3

Only way to make 33 with five digits is 3/6/7/8/9. This implies that the 3 of that "33" shape should go in the middle-top box, meaning the 8 and 9 must be left somewhere in the "33" part of the top-left box, e.g. it could be

36X XXX XXX

178 3XX XXX

AX9 XXX XXX

However, the only way to make the two-digit 17 which adjoins with the centre-left box below is using 8 and 9, which seems to me to mean that you have to use the 8 or 9 twice in the same 3x3 box! In other words, "A" (above) MUST be 8 or 9, yet it's already been used in that box.

Something's wrong, or am I just completely stupid!???