On page 11 in Electrodynamics of Continuous Media by L.D. Landau & E.M. Lifshitz, the authors describe the method of inversion: Laplace's equation in spherical coordinates is unaltered if $r$ is replaced by $r'$ and $\phi$ is replaced by $\phi'$ such that:
$$r = R^2/r',\,\phi = r'\phi'/R$$ Here, R is some constant having the dimensions of length. Thus, if $\phi(\textbf{r})$ satisfies Laplace's equation, then so does the function $\phi'(\textbf{r}')=R\phi(R^2\textbf{r}/r'^2)/r'$. How does one prove this?
I'll add my attempt below as an answer if someone via the comments can help me correct it; otherwise any answers would be appreciated:
Laplace's equation in Spherical coordinates is
$$\nabla^2 \phi = \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \phi}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2\phi}{\partial \psi^2}=0$$
which simplifies to
$$\nabla^2 \phi = \frac{\partial }{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \phi}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\phi}{\partial \psi^2}=0$$
Only the first term depends upon r: $$\frac{\partial }{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)= \frac{dr'}{dr}\frac{\partial}{\partial r'}\left(r^2\frac{dr'}{dr} \frac{\partial \phi}{\partial r'}\right)= -\frac{R^2}{r^2}\frac{\partial}{\partial r'}\left(-r^2\frac{R^2}{r^2} \frac{\partial \phi}{\partial r'}\right)= r'^2\frac{\partial^2 \phi}{\partial r'^2} $$
Replacing $\phi$ with $r'\phi'/R$:
$$\begin{align} &= \frac{r'^2}{R}\frac{\partial}{\partial r'}\left(\phi'+ \frac{r'\partial \phi'}{\partial r'}\right)= \frac{r'^2}{R}\left(2\frac{\partial\phi'}{\partial r'}+ \frac{r'\partial^2 \phi'} {\partial r'^2}\right)= \frac{r'}{R}\frac{\partial }{\partial r'}\left(r'^2 \frac{\partial \phi'}{\partial r'}\right)\\&\neq \frac{\partial }{\partial r'}\left(r'^2 \frac{\partial \phi'}{\partial r'}\right) \end{align} $$
Almost there, but not quite.
You have everything right, you just have apply the transformation to $\phi$ in the rest of Laplace's equation as well. Rewriting Laplace's equation as you have it, $$ \frac{\partial }{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)= -\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \phi}{\partial\theta}\right) - \frac{1}{\sin^2\theta}\frac{\partial^2\phi}{\partial \psi^2}. $$ Then, applying your result for the left hand side and expressing the right hand side also in terms of $\phi'$ gives $$ \frac{r'}{R}\left[\frac{\partial }{\partial r'}\left(r'^2 \frac{\partial \phi'}{\partial r'}\right)\right] = \frac{r'}{R}\left[ -\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \phi'}{\partial\theta}\right) - \frac{1}{\sin^2\theta}\frac{\partial^2\phi'}{\partial \psi^2}\right]. $$ Cancelling the prefactor of $r'/R$ on both sides then shows that $\phi'$ is also a solution of Laplace's equation.