I think that I am not fully understanding what the exercise asks me to show. Here it is:
(Fulton, Algebraic Curves, 2008)
By definition, a variety $X$ is an open subset of $V$, where $V$ is an irreducible algebraic subset. Here $\mathcal{O}_p(X)=\mathcal{O}_p(V)$ for every $p\in X$, and $\varphi$ is just the function induced by $f$.
But, unless I have misunderstood, what the exercise asks me to show is false. If we fix $\lambda \in k$, then the pole set of $z=\frac{1}{f-\lambda}$ is the set of all $P\in V$ such that $f(P)=\lambda$. In order to prove this, let $Q$ be the pole set of $z$ and let's show that $Q=f^{-1}(\lambda)$. It is obvious that $Q\subseteq f^{-1}(\lambda)$. To prove the reverse inclusion, we must show that if $z=\frac{a}{b}$ where $a,b\in \Gamma(V)$, then $b(v)=0$ for every $v\in f^{-1}(\lambda)$. But this is easy, because $b=(f-\lambda)a$.
Ok, so $\varphi^{-1}(\lambda)=f^{-1}(\lambda)\cap X$ is missing all the elements in the pole set of $z$ which do not lie in $X$, am I right? Of course, maybe it is possible to show that all the elements in the pole set of $f$ lie in $X$, but that seems quite impossible. I mean; it would be almost the same as proving that the pole set of any element of $\Gamma(V)$ which can be written as $\frac{1}{a}$ is a subset of every open subset of $V$, which I do not expect to be true.
Maybe we have to change the definition of pole set when we are dealing with open subsets of irreducible algebraic sets?