What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$? 
Let $B_a$ be the open ball{$x: |x|^2<a$} in $\mathbb{R}^k, |x|^2 = \sum x_i^2$.
  What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$?

Here I want some justification to equate $|x|^2$ with $x^2$, I don't know to justify; but otherwise I don't know how to solve it.
 A: Try to think a minute about what this function does; as Rahul Narain stated, it multiplies a vector by a "scalar" which depends on the norm of the vector.
Second, remark the spherical invariance of this scalar; any vector with norm $r$  (origin to any point on the sphere of radius $r$) will have an associated scalar $\frac{a}{\sqrt{a^2 - r^2}}$.
Third, note that this scalar tends to $1$ when the radius tends to $0$, and that it tends to infinity when the radius tends to $a$. In other words, it "maps" the open $a$-ball to $\mathbb{R}^n$.
Now the only inverse you have to find is that of that last sentence... :) (Well, you do have to understand that mapping as well ;).

Ok, I gave you some time to think about it, now let's see the answer. As we said it's a mapping from $\mathbb{R}^n$ to the open $a$-ball, the question that drives us crazy is which one?
Take a look at the norm of the resulting vector. For any vector of $\mathbb{R}^n$ (this includes any vector in any open ball), we can write $\vec{x} = r\vec{u}$, where $\vec{u}$ is some unit vector, and $r$ is the norm of $\vec{x}$.
The norm of the "image" of $\vec{x}$ by our function (say $\vec{y}$) is $\frac{ar}{\sqrt{a^2 - r^2}}$. Because our application preserves the orientation, we can also write $\vec{y} = s\vec{u}$ (where $\vec{u}$ is the same than before), and we have $|y|=s=\frac{ar}{\sqrt{a^2 - r^2}}$. 
If we simply inverse this last relation (see below) we obtain the inverse function, that is
$
\frac{a s}{\sqrt{a^2 + s^2}} \vec{u} = \frac{a y}{\sqrt{a^2 + |y|^2}}
$.

Details about the inversion:
$$
s^2 (a^2 - r^2) = a^2r^2 \iff a^2s^2 = r^2(s^2+a^2) \iff r = \frac{as}{\sqrt{a^2+s^2}}
$$
