What is the number of arrangements in the word "DAUGHTER" where vowels are never together?

You should have applied the Inclusion-Exclusion Principle rather than the Complement Principle.

Since the eight letters of DAUGHTER are distinct, they can be arranged in $8!$ ways. From these, we must exclude those arrangements in which a pair of vowels is consecutive.

A pair of consecutive vowels: There are $\binom{3}{2}$ ways to choose two of the three vowels to be in the pair. If we treat the pair of vowels as a single object, we have seven objects to arrange, the pair of vowels and the other six letters. Since the objects are distinct, they can be arranged in $7!$ ways. Within the pair of vowels, the vowels can be arranged in $2!$ orders. Hence, the number of arrangements of DAUGHTER in which there is a pair of consecutive vowels is $$\binom{3}{2}7!2!$$

If we subtract these arrangements from the total, we will have subtracted too much. For instance, consider the arrangement DAEUGHRT. We subtract it once when we designate AE as the pair of consecutive vowels and once when we designate EU as the pair of consecutive vowels. Thus, we have subtracted each arrangement with two pairs of consecutive vowels (which, in this case, means three consecutive vowels) twice, once when we designate the first two vowels as the pair of consecutive vowels and once when we designate the last two vowels as the pair of consecutive vowels. We only want to subtract such arrangements once, so we must add them back.

Two pairs of consecutive vowels: This means that the three vowels must be consecutive. If we treat the block of three vowels as a single object, we have six objects to arrange. Since they are distinct, the objects can be arranged in $6!$ ways. Within the block, the vowels can be arranged in $3!$ orders. Hence, there are $$\binom{3}{3}6!3!$$ such arrangements.

By the Inclusion-Exclusion Principle, there are $$8! - \binom{3}{2}7!2! + \binom{3}{3}6!3!$$ arrangements of the letters of the word DAUGHTER in which no two vowels are consecutive.

I believe you are wrong because you are double counting when you get the same word with $3$ vowels next to each other AUE but at one time you counted the UE as the same letter while another with AU as the same letter.

I will try a different approach. First pick the 5 consonants and permute them. There are $5!$ ways to do this. Then place the vowels in between them. There are 6 places for them so there are $6!/3!$ ways to place them. Total amount is $5!6!/3!$

More than a hint...Arrange the consonants amongst themselves in $5!$ ways and then independently insert the three vowels into the six spaces available (between the consonants and at the beginning and end)in $\frac{6!}{3!}$ ways and multiply together.