Consider $y = f(x)$ where $x \in I$ for some interval $I$. The typical parametrization goes set $x = t$ and so $y = f(t)$ and then we get a parametrization $r(t) = (t,f(t))$. Things to point out here is that you are parametrizing the graph of $f$ i.e if $f: I \to \mathbb{R}$ then $r(t) \in \mathbb{R}^2$. In your case $f: \mathbb{R}^2 \to \mathbb{R}^2$; hence your paramterization (if it exists) should be in $\mathbb{R}^4$. Here $z = f(x,y)$ is determined by $x,y$ i.e points on the graph of $f$ are of the form $(x,y, x^2y, yx)$. The problem here is that there is no mentioned relationship between $x,y$ so you have no chance of giving a parametrization. For instance, if you knew $y = g(x)$ then you would set $x =h(t) \Rightarrow y = g(h(t))$ and so you could get the parametrization:
$$r(t) = (h(t),g(h(t)),h(t)^2g(h(t)),g(h(t))h(t))$$