Here is a solution inspired by Beastly Gerbil's observation in a comment that:

>! If we label the numbers A through I in reading order, then (A+B-C)+(D+E+F)+(G+H+I) is 12+12+11=35 by the conditions on the rows; on the other hand, A+B+C+D+E+F+G+H+I=45, the sum of all 9 values. Subtracting the first equation from the second, we get 2C=10, or C=5.

From here, we can finish with almost no casework:

>! The first row now tells us A+B-5=12, or A+B=17, so A and B are 8 and 9 in some order. Setting B=8 means that E×H=37-8=29 in the second column, which is impossible, so A=8 and B=9. Then E×H=28, which means E and H are 4 and 7 in some order.

>! From the first column, 8×D-G=23. D must be at least 3 for 8×D to be bigger than 23, but if D is 4 or more, then G=8×D-23 is too big. So D=3 and G=1. 

>! If E=4 and H=7, then G+H+I=11 means I=4, which was already used. So E=7 and H=4. G+H+I=11 gives I=6 and D+E+F=12 gives F=2, so we have the complete solution: A=8, B=9, C=5, D=3, E=7, F=2, G=1, H=4, I=6.

We didn't even use the condition in the third column, though we can check that it's satisfied at the end.