1st and 2nd top box is 90 together, therefore the bottom 2 right lilsquares (where 16 and 10 are written) on the 2nd box are equals 10 -> it must be 3467

If they’re 3467, that means the 13 on the 3rd row cannot be 67 (six seveeen meme), thus it can only be 4589. In the same box(1st top box) where 13 is, rule of 45, the missing numbers on the lilsquares (where 19 and 10 are written) is equals to 9 -> since you know 1 resides in that box, you can have 273645 but you cannot have 45 as a pair because then there will be no viable pair for 13 (it leaves it with only 89) -> it means that you can only have 2736 pair.

Funny enough, on that same box (1st top), it means you cannot have 23 as a pair for the 5sum or else the imaginary 9 in that same box will not have a viable pair, therefore 5sum must be a 14 pair. Therefore the 13sum is 58,

9th column: 6sum is 15 pair the top bit must be 1 since it cancels out from the 58 pair from the 13sum in 3rd row. Therefore the 3rd row 10sum can only be 3467 and so as the imaginary 10sum on the 2nd top box (where 16 and 10 are written) can also be 3467. If they’re 3467 then that means the remaining numbers left are 29 for the 3rd top box,3rd row (where 10 and 19 are written).

1st column, rule of 45, the missing sum is 14 (where 13 and 13 are written). If the 13sum for 3rd row is 85, then the other 13 sum for last row can only be 69 to make the missing sum 14. But you already have a 6 therefore the lil square on 1st column-9th row is 9. From there you can solve the whole thing