Preimage of two-dimensional function We have the function $f:(\mathbb{R}^2,\|\cdot\|_2)\rightarrow (\mathbb{R},|\cdot |)$ with \begin{equation*}f(x,y)=\begin{cases}y-x & y\geq x^2 \\ 0 & y<x^2\end{cases}\end{equation*} 
How could we draw the preimage $f^{-1}((1,\infty))$ ? Could you give me a hint? 
 A: Sice $0\notin (1,\infty)$, you can focus on the first part and intersect the appropriate half space with that region there.
A: Since we want the pre-image of $(1,\infty)$, we can ignore all regions where the function is equal to $0$. That is, focus on $y \ge x^2$. In particular, we have $f(x,y) = 1 \iff y - x = 1 \iff y = x + 1$. Hence, $f(x,y) > 1 \iff y > x + 1$ and $y \ge x^2$. 
At this point, we could stop here and say that you can draw the graphs of $y = x + 1$ and $y = x^2$, and shade the regions that are above both lines (including only the line of the second graph).
To be even more explicit, notice that $x^2 = x + 1 \Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$. A quick test shows that $x^2 > x + 1$ for $x < \frac{1 - \sqrt{5}}{2}$ or $x > \frac{1 + \sqrt{5}}{2}$, and $x^2 < x + 1$ for $\frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2}$. Hence, 
$$f^{-1}((1,\infty)) = \{(x,y) \mid y \ge x^2 \text{ and } x < \frac{1 - \sqrt{5}}{2} \text{ or } x > \frac{1 + \sqrt{5}}{2}\} \cup \{(x,y) \mid y > x + 1 \text{ and } \frac{1 - \sqrt{5}}{2} \le x \le \frac{1 + \sqrt{5}}{2}\}$$
