This is a problem out of Guilleman and Pollack, page 5.
Okay, intuitively I see what this does to points in an open ball of radius $a$, $B_a$. The ball gets mapped onto $\mathbb{R}^k$. $\frac{a}{\sqrt{a^2 - |x|^2}}$ is the cosecant of the angle between $x$ and a vector at the intersection of the hyperplane normal to $x$ with $\partial B_a$.
I cannot find the inverse map analytically but maybe I can argue geometrically. Given $y \in \mathbb{R}^k$, its inverse image with be in the direction of $y$, $\frac{y}{|y|}$. Now, the magnitude will be some sort of scaling of $|y|$. I am assuming the scaling is $\frac{\sqrt{a^2-|y|^2}}{a}$, so that the inverse map is $y \mapsto \frac{\sqrt{a^2-|y|^2} y}{a} \in B_a$. Is this the right thought process?
Well, you are on the right track and almost got it. Define the scalar function $$\kappa(x) = \frac{a}{\sqrt{a^2 - |x|^2}}.$$ Then your map looks like this $x \, \mapsto \, y = \kappa(x) \, x$, which means that $x$ and $y$ are collinear with the origin $0$, i.e. they are linearly dependent as vectors. Fix a $x$ and let $y = \kappa(x) \, x$. Then there exists a vector $\hat{x}$ such that $|\hat{x}| = a$ and real numbers $\lambda < 1$ and $\mu$ such that $x = \lambda \, \hat{x}$ and $y = \mu \, \hat{x}$. Then $$\mu \, \hat{x} = y = \kappa(x) \, x = \frac{a}{\sqrt{a^2 - |x|^2}} \,\, x = \frac{a \, \lambda}{\sqrt{a^2 - \lambda^2 |\hat{x}|^2}} \,\, \hat{x} = \frac{a \, \lambda}{\sqrt{a^2 - \lambda^2 a^2}} \,\, \hat{x} = \frac{ \lambda}{\sqrt{1 - \lambda^2}} \,\, \hat{x}$$ which is possible exactly when $$\mu = \frac{ \lambda}{\sqrt{1 - \lambda^2}}.$$ Solve for $\lambda$ and get $$\lambda = \frac{ \mu}{\sqrt{1 + \mu^2}}$$ Now going backwards \begin{align} x &= \lambda \, \hat{x} = \frac{ \mu}{\sqrt{1 + \mu^2}} \, \hat{x} = \frac{ y}{\sqrt{1 + \mu^2}} = \frac{ y}{\sqrt{1 + \mu^2 \frac{|\hat{x}|^2}{a^2}}} = \frac{a \, y}{\sqrt{a^2 + \mu^2 |\hat{x}|^2}} \\ &= \frac{a \, y}{\sqrt{a^2 + |\mu \, \hat{x}|^2}} = \frac{a}{\sqrt{a^2 + |y|^2}} \,\, y \end{align} so your inverse looks like this $$x = \frac{a}{\sqrt{a^2 + |y|^2}} \,\, y$$