You have two equals signs there. Did you mean a plus sign in there somewhere?
Different people learn this method different ways. I learned it this way:
If we follow your substitution of then and (This is how I "find" . If you wish, you can write .) Plugging into the DE yields
Separating out yields
(absorbing the minus sign into the constant).
At this point, as with all solutions to any differential equation, you should verify that this implicitly defined function satisfies the original DE. The reason to differentiate now is that if you do, the constant will go away, which is what you want when you're verifying the solution to a DE. The solution I've got here is, incidentally, the same as yours, I believe. So far, so good. If you do want to isolate then take the of both sides, and then multiply the resulting equation by to obtain
On what interval(s) is this solution valid?