> 
> Let $k$ be a algebraicly closed field and $p \in k^2$.
> 
> 
> It is to show, that if $z \in k(k^2)=\mathrm{Frac}(k[X,Y])$ is defined
>  in every point $q \in k^2\backslash \{p\}$, then it already holds that
>  $z \in k[X,Y]$.
> 
> 
> 

($z\in k[X,Y]$ defined in $q \in k^2$ means that $\exists u,v \in k[X,Y]$ with $v(q)\neq 0$ and $z=\frac{u}{v}$.)

I started with saying that $k^2 \backslash \{p\}\subseteq \mathrm{Def}(z)=\{q \in k^2 |\,\, \mathrm{z \,\, defined \,\, in \,\,} q\}$ and so it has to hold that $p \in \mathrm{Def}(z)$, because $z \in k[X,Y]$ if and only if $\mathrm{Pol}(z):=k^2 \backslash \mathrm{Def}(z)=\emptyset$ (something that was already proven). But that's as far as I got.