How many words can be written with using the letters ELLHELY so that 3 L's are never in a row and no E's are ever next to each other? My approach was as follows: 
1: Find the number of total different words that can be made. Did this by $\frac{7!}{3!2!}$ since three L's repeat and two E's repeat. Answer: 420.
2: Find the number of cases where all three L's are in a row. Did this by $\frac{5!}{2!}$ where I treat the three L's as one item, thus 7 becomes 5 and only the 2 from the repeating E's is left on the bottom. Answer: 60.
3: Find the number of cases where two E's are next to each other. I treat the E's as one item and use $\frac{6!}{3!}$ to find this where the 3 on the bottom represents the repeating L's. Answer: 120.
4: I subtract the amount of "invalid" cases which we found in 2 and 3 from the total that we found in 1. So 420 - 30 - 20 = 370. Answer 370 ways.
I know this would be correct if these were separate questions, so is it still correct when I apply both of them (step 4)?
Thank you!
Edit: Counted the factorials wrong, fixed now.

You forgot something: You substracted the words, where both conditions (the one on the $\sf L$s and the one on the $\sf E$s) are met, twice. So you have to add them again, once. There are $4!$ of them (we just treat the $\sf L$s and the $\sf E$s as one letter each), so alltogether we have
$$ \frac{7!}{3!2!} - \frac{5!}{2!} - \frac{6!}{3!} + 4! 
  = 420 - {\color{red}{60}} - {\color{red}{120}} + 24 = 264 $$
