Actually both of the formulas you have are wrong in some way. In both, the $\forall y$ only binds the $y$ in $P(y)$, leaving the $y$ in $L(x,y)$ free. In your professor's sentence, the parenthesis before $\forall y$ should come right after it. Finally, there's no need for the outer brackets. Thus:
\neg\exists x~\Big(P(x) \land \forall y \big(P(y) \to L(x,y)\big)\Big)
However, this isn't right: it's equivalent to
\forall x~\Big(P(x) \to \neg\forall y~\big(P(y) \to L(x,y)\big)\Big) &\iff \forall x~\Big(P(x) \to \exists y~\big(P(y) \land \neg L(x,y)\big)\Big)
which means "everybody doesn't love someone", or equivalently, nobody loves everybody. Everybody does not mean anybody. I am assuming that "nobody loves anybody" does not mean the same thing as "nobody loves just anybody" (i.e. everybody).
I take "nobody loves anybody" to mean for all people $x$ and $y$, $x$ does not love $y$:
\forall x~\forall y~\Big(\big(P(x)\land P(y)\big)\to \neg L(x,y)\Big)