Combination Problem: Arranging letters of word DAUGHTER

I don't know what possibility you missed out, but I'd think it's easier to calculate how many possibilities there are where all the vowels do appear together, and subtract that from the total $8!=40\,320$.

If all the vowels appear together, then we can think of those three together as one letter, which means there are $6!$ different words we can write. But for each of those there are $3!$ ways of rearranging the vowels within the block. So there are $6!\cdot 3! = 4\,320$ different words where all the vowels appear together.

This means there are $40\,320 - 4\,320 = 36\,000$ words where all the vowels do not appear together.

You haven't included the possibility of permuting e.g. the U and E in UE when they're together (and the same in the other two cases).

When this is included, your method works:

$$5! \times \left(\binom{6}{3} \times 3!+3 \times \binom{6}{2} \times 2! \times 2! \right)=36000.$$