I need to solve using the method of separation of variables. I got to this point but I'm not sure if I'm on the right track:
$$\frac{-1}{2e^{2y}} + C = \frac{e^{3x}}{3} + C$$
It looks like now I need to just solve for $y$ but how do I get rid of the $e$'s?
You're on the right track. As mentioned in the comments you don't need a "$+C$" on both sides of the equation. Each integral contributes an additive constant, the two of which we can combine into a single constant.
Now you just need to solve for $y$. Start by isolating the only term with a $y$ in it.
$$\frac{-1}{2} e^{-2y} = \frac{1}{3}e^{3x}+C$$
$$ e^{-2y} = -\frac{2}{3}e^{3x}-2C$$
$$ e^{-2y} = -\frac{2}{3}e^{3x}+C$$
Notice that I just absorbed the $-2$ into a new definition of $C$.
Now we need to use logarithms to get rid of the $e$ on the left.
$$ -2y = \ln( -\frac{2}{3} e^{3x} + C) $$
$$ y = -\frac{1}{2}\ln( -\frac{2}{3} e^{3x} + C) .$$
And we are done. Each possible value of $C$ will correspond to a different solution to the differential equation with different initial data.