Let $\Omega \subset \mathbb{R^3-\{0\}}$ and $u:\Omega \rightarrow \mathbb{R}$ harmonic in $\Omega$. Show that $$v(x^1,x^2,x^3):=\dfrac{1}{|x|}u\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)$$ is harmonic in the region $\Omega' :=\{x\in \mathbb{R^3}:\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)\in \Omega\}$
a. Is there a "deeper" reason for this?
b. Is there an analogous result for arbitrary dimension $d$?
My solution for the first part is the following:
Because $\Delta(fg)=\Delta(f)g+f\Delta(g)+2\nabla f \nabla g$ then
$$\Delta v(x^1,x^2,x^3)=\Delta\Big(\dfrac{1}{|x|}\Big)u\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)+\dfrac{1}{|x|}\Delta u\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)+2\nabla\Big(\dfrac{1}{|x|}\Big)\nabla u\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)=2\nabla\Big(\dfrac{1}{|x|}\Big)\nabla u\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)$$ for $x'\in \Omega$. With $x'=\Big(\dfrac{x^1}{|x|^2},\dfrac{x^2}{|x|^2},\dfrac{x^3}{|x|^2}\Big)$
Because $u$ is harmonic in $\Omega$ and $\dfrac{1}{|x|}$ is harmonic in $\mathbb{R^3}-\{0\}$.
What can I do with the last term? (I assume that those vectors are orthogonal but don't know how to show it.)
Also I don't know what to do with questions a and b. Thanks in advance for any hints.