What you did so far is correct, and you only have to fill in the "..."s in your subproofs.
The second subproof is easy: You already got your conclusion $\forall x \neg P(x)$ as an assumption -- just reiterate (R) that line and you're done.
As for the first subproof, the idea is to derive a contradiction between $\neg P(a)$ for some individual $a$ in the second premise -- of which we know there is an instance, since we are given $\forall x Q(x)$ -- and the universal claim $\forall x P(x)$:
Assume $Q(a) \to \neg P(a)$ for an arbitrary individual $a$. By the third premise $\forall x Q(x)$ we know by universal instantiation ($\forall\!$ E) that $Q$ indeed holds of $a$. By modus ponens ($\to\!$ E), we can conclude $\neg P(a)$. But this contradicts the proposition that $P(a)$, which we get out of the assumption $\forall x P(x)$ by $\forall\!$ E, so we get a contradiction $\bot$. From this contradiction we are allowed to conclude anything thanks to ex falso quodlibet ($\bot$) -- conveniently, we can choose $\forall x \neg P(x)$ as the next conclusion. Since we were able to derive this conclusion under the assumption that $Q(a) \to \neg P(a)$ holds for some individual $a$, and by the second premise we know that at least one such individual does exist, we may apply $\exists\!$ E on the existential formula and the subproof, and thereby discharge the assumption $Q(a) \to \neg P(a)$ to conclude $\forall x \neg P(x)$ for sure.