Let $X =\{ x_1x_4=x_2x_3\;, (x_2,x_4) \neq (0,0)\} \subset \mathbb{C^4}$, i.e. not both of $x_2,x_4$ are zero. Define a function $\phi$ on $X$ by $\phi(x)=\left\{\begin{matrix} \frac{x_1}{x_2} & ,x_2 \neq 0 \\ \frac{x_3}{x_4}& ,x_4 \neq 0 \end{matrix}\right.$.
Prove that there are no two polynomials $f,g \in \mathbb{C}[x_1,x_2,x_3,x_4]$ such that for all $x \in X$, $\phi(x)=\frac{f(x)}{g(x)}$.
In fancier words, $\phi$ is a regular function on an open subset $V(x_1x_4-x_2x_3) \setminus V(x_2,x_4)$ of an affine variety that can not be defined by a single choice of polynomials.
I don't have a good approach. I tried the following: we know that on the open subset where $x_2\neq0$ and $x_4\neq0$ we have pointwise both $f(x)x_2=g(x)x_1$ and $f(x)x_4=g(x)x_3$, and I think these relations hold on all of $X$, since they are true on a Zariski-dense subset of $X$, and the set where they hold is a closed subset of $X$. This tells us a bit about the nature of such $f,g$ since e.g. $g$ must vanish whenever $x_2=0,x_1 \neq 0$.