The first thing to notice is that the the values of Y-ehe are multiples of the corresponding row. 45 is a multiple of 9, 40 a multiple of 8 and 14 a multiple of 7, etc.
The other thing is that the X-akseli and Y-ehe values sum up to the same value, 194.
If you look at the matrix, you see that the multiples in Y-ehe almost match the number of blank cells. There are 2 empty cells in row 7 and Y-ehe(7) is 2x7. So if you fill the blank cells with the row number, the cells in a row nicely sum up to the Y-ehe value.
Naturally you will want to compute the column sums, and you discover the X-akseli values.
Note however that you have to remove the initial 6 to make it work. Probably a typo.
All this teaches us one thing: the value of a cell is the row number.
Now I will boldly assume, out of nowhere, that the goal is to go from the top-left to the bottom-right corner by minimizing the total cell count. Here are 2 options:
The number at the bottom-right is the sum of all cells. As you can see, the second path, with a score of 80, is better than the first one, with a score of 82. It is longer but uses smaller cell values. There are other paths, but they result still larger scores. The solution on the right shows the optimal path.
I have no explanation about the mismatched sums or where the hint "Think in all 3 dimensions" comes into play.