Does there exist an inverse of the following function with given domain?
$$f(x,y)=\left( \frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right), \quad (x,y) \in \mathbb{R}^2$$
$$(\mathbb{R}^2= \{ (x,y):x,y \text{ are real numbers}, \text{ excluding } (x,y)=(0,0) \})$$
I know when the function is of a single variable its inverse can be visualised flipping around the $x=f(x)$ line in the $(x,f(x))$ plane, but how would you interpret the inverse of a function as above (more dimensions)?
If $g(u,v) = (h(u,v), s(u,v))$ is the inverse function, then $(f\circ g)(u,v) = (g\circ f)(u,v) = (u,v).$ That gives you a system of equation: $$\begin{cases}\dfrac{h}{h^2+s^2} = u& \\ \dfrac{s}{h^2+s^2} = v\end{cases}.$$
When you solve this for $h$ and $s$ (which is a simple algebra basically), you get a formula for $g(u,v) = \left(\dfrac{u}{u^2+v^2}, \dfrac{v}{u^2+v^2}\right)$
$$ f^{-1}(u, v) = \left( \frac{u}{u^2 + v^2}, \frac{v}{u^2 + v^2} \right). $$ As for the graphical interpretation, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a function from the $x$-$y$ plane to the $u$-$v$ plane, then the graph of $(x, y, f^{-1}(x, y))$ (in $\mathbb{R}^4$) is produced by flipping that of $(x, y, f(x, y))$ (in $\mathbb{R}^4$) with respect to the plane $x = u, y = v$.
