Set-up. Let $k$ be an algebraically closed field.
Let $V\subseteq \mathbb{A}^n(k)$ be an irreducible Zariski-closed subset.
Let $f$ be a rational function on $V$ (in other words, let $f$ be an element of $k(V)$).
Question. I want to show that the set $$W:=\{P \in V \, : \, f \text{ is regular at } P \;\text{ and }\; f(P)=0\}$$ is a Zariski-closed subset of $V.$
Attempt. We can write $f=F/G\,$ for some $F,G \in k[V].$
Then, for any $P \in V$ at which $f$ is regular, we have $$f(P)=0 \Leftrightarrow F(P)=0$$ and so $W=\mathbb{V}(F),$ the zero locus of $F$ in $V,$ which is Zariski-closed by definition.
EDIT. It has been pointed out below that the correct statement is that
"$W$ is a Zariski-closed subset, not of $V,\,$ but of $\;\mathrm{dom}(f):=\{P \in V \, : \, f \text{ is regular at } P\}$".