Title says it all.
I tried defining $g(x) = |x|^2$ and $h(x) = (|x|^2)^{\frac{-n}{2}}$
Now $f(x) = g(x) \cdot h(x)$ and so $\Delta f=g\Delta h+2\langle\nabla g,\nabla h\rangle + h \Delta g$
But calculating $\nabla h$ and more so $\Delta h$ seems like a really un-fun thing to do. There should be an easier way, maybe with chain rule or something.
From my calculations, $\nabla g = \begin{pmatrix}2x_1\\\vdots\\2x_n\end{pmatrix}, \Delta g=2n, \nabla h = \begin{pmatrix}\frac{-nx_1}{|x|^{n+2} \\ \vdots \\ \frac{-nx_n}{|x|^{n+2}}}\end{pmatrix}$
My patience broke down trying to compute the laplacian of $h$.
Edit: Incase it wasn't clear, $f: \mathbb R^{n} \setminus \{0\} \to \mathbb R, |x| = \sqrt{\sum_{i=1}^{n}x_i^2}$
It is easier to first show that, if $u(x)=u(|x|)$ is a function of $|x|$ only, then $$\Delta u(x)=u''(|x|)+\frac{n-1}{|x|}u'(|x|).$$