From my PDE class:
suppose that $\Delta u = 0$ for $u(x_1,\dots,x_d)$. Then show $$\Delta\left[|x|^{2-d}u\left(\frac{x}{|x|^2}\right)\right] = 0$$
Furthermore when $d=2$, $\Delta$ is invariant under conformal transformations.
I have tried expanding this using the identity $\Delta(\phi\psi) = \phi\Delta\psi + \psi\Delta\phi + 2 \nabla\psi\cdot\nabla\phi$. I have managed to compute
$$\nabla\left[|x|^{2-d}\right] = x(2-d)|x|^{-d}$$ $$\Delta\left[|x|^{2-d}\right] = d(2-d)|x|^{-d}$$
But the expressions for the other term turn out quite messy
$$\nabla\left[u(y)\right] = -\frac{2x(x\cdot\nabla u)(y)}{|x|^4} + \frac{1}{|x|^2}(\nabla u) (y)$$ $$\Delta\left[u(y)\right] = -\frac{2d(x\cdot\nabla u)(y)}{|x|^4} - \frac{2}{|x|^6} ((x\cdot\nabla)(\nabla u))(y)\cdot x$$
Where $y=\frac{x}{|x|^2}$. I'm not sure if this is the right way to go about proving this idenetity or if the above expressions are even correct.
For the second statement in the case where $d=2$ the above identity becomes
$$\Delta\left[u\left(\frac{x}{|x|^2}\right)\right] = 0$$
If we consider $x=z$ as a complex number we have
$$\Delta\left[u\left(\bar z^{-1}\right)\right] = 0$$
Since any conformal mapping is locally a mobius transform. I think this means that combining this property and invariance of $\Delta$ under scaling/translation, we get an invariance under conformal mapping in general, but I'm not sure how to argue this.