Prove that $v(x)=\frac{1}{|x|^{n-2}}u(\frac{x}{|x|^2})$ is harmonic 
Let $\Omega$ be an open set in $\mathbb{R}^n$ ($n\geq 3$), and $u:\Omega \to \mathbb{R}$ be a harmonic function. Let 
  $$\Omega'=\Big \{x:\frac{x}{|x|^2}\in \Omega \Big \} \mathrm{\ \ and\ \ }v(x)=\frac{1}{|x|^{n-2}}u\left(\frac{x}{|x|^2}\right)$$ 
  Prove that $v(x)$ is harmonic in $\Omega'$.

My attempt:
We compute that 
$$\frac{\partial}{\partial x_i}u\left(\frac{x}{|x|^2}\right)
=\frac{1}{|x|^4}\left(|x|^2u_i-2x_i\sum_{k=1}^nu_kx_k\right)$$
$$\frac{\partial}{\partial x_i}\left(\frac{1}{|x|^{n-2}}\right)
=\frac{(2-n)x_i}{|x|^n}$$
and $$\frac{\partial^2}{\partial x_i^2}\left(\frac{1}{|x|^{n-2}}\right)=(2-n)\left(1-\frac{nx_i^2}{|x|^2}\right)$$
Therefore $\displaystyle \Delta \left(\frac{1}{|x|^{n-2}}\right)=0$
Thus, we have
\begin{align*}
\Delta v(x)&= \left(\frac{1}{|x|^{n-2}}\right)\Delta u\left(\frac{x}{|x|^2}\right)
+u\left(\frac{x}{|x|^2}\right) \Delta \left(\frac{1}{|x|^{n-2}}\right)
+\nabla\left(\frac{1}{|x|^{n-2}}\right)\cdot \nabla u\left(\frac{x}{|x|^2}\right) \\
&=\nabla\left(\frac{1}{|x|^{n-2}}\right)\cdot \nabla u\left(\frac{x}{|x|^2}\right) \\
&=\sum_{i=1}^n \frac{(2-n)x_i}{|x|^n}\cdot \frac{1}{|x|^4}\left(|x|^2u_i-2x_i\sum_{k=1}^nu_kx_k\right) \\
&=\frac{n-2}{|x|^{n+2}}(u_1x_1+u_2x_2+\cdots +u_nx_n)
\end{align*}
What's wrong with my calculation?
 A: Here's the mistake: 
$$
\Delta v(x) 
= \left(\frac{1}{|x|^{n-2}}\right)\Delta \left( u\left(\frac{x}{|x|^2}\right) \right)
+u\left(\frac{x}{|x|^2}\right) \Delta \left(\frac{1}{|x|^{n-2}}\right)
+\nabla\left(\frac{1}{|x|^{n-2}}\right)\cdot \nabla \left( u\left(\frac{x}{|x|^2}\right) \right)
$$
You're right about the second term (it disappears) and the third term (you look at the gradient of the composition), but instead of $\Delta \left( u\left(x/|x|^2\right) \right)$ you're looking at $(\Delta u ) \left(x/|x|^2\right)$. 
Derivatives of compositions can be tricky when it comes to notation. 

All computations involved are quite straightforward, for example 
$$
\Delta (u \circ f) = \sum_{ijk} \partial_{jk} u \cdot \partial_i f^j \cdot \partial_i f^k + \nabla u \cdot \Delta f.
$$
In our case $f(x) = x/|x|^2$, so one can compute 
\begin{align*}
\sum_{i} \partial_i f^j \cdot \partial_i f^k & = -2 |x|^{-4} \delta_{jk}, \\
\Delta f & = (4-2n) |x|^{-4} x 
\end{align*}
(if I'm not mistaken). Thanks to $\Delta u = 0$, the first term has no contribution, while the other cancels out with $\nabla\left(|x|^{2-n}\right)\cdot \nabla \left( u \circ f \right)$. 
