I know how to prove $$\nabla^2(\frac{1}{r}) = -4\pi\delta^3(\textbf{r}),$$ where in Spherical coordinates, $$ \nabla^2 = {\frac{1}{r^2} \, \frac{\partial}{\partial r} \left( r^2 \, \frac{\partial}{\partial r} \right)} + \frac{1}{r^2 \sin\theta} \, \frac{\partial}{\partial\theta} \left( \sin\theta \,\frac{\partial}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta} \, \frac{\partial^2}{\partial\phi^2}.$$
Similarly when I try to prove $$\nabla^2(\frac{1}{R}) = -4\pi\delta^3(\textbf{R}),$$ where $$R=|\textbf{R}|=|\textbf{r} - \textbf{r}'|,$$ I need to write down the Laplacian $\nabla^2$ in terms of $R$. How can I explicitly show this?
Note the following
\begin{eqnarray*} \partial _{\mathbf{x}}\frac{1}{R} &=&\partial _{\mathbf{x}}\frac{1}{|\mathbf{ x-y}|} \\ \frac{\partial }{\partial _{x_{k}}}\frac{1}{|\mathbf{x-y}|} &=&\frac{ \partial (x_{k}-y_{k})}{\partial _{x_{k}}}\frac{\partial }{\partial (x_{k}-y_{k})}\frac{1}{|\mathbf{x-y}|}=\frac{\partial }{\partial R_{k}}\frac{ 1}{R} \\ \frac{\partial }{\partial _{x_{k}}}\frac{\partial }{\partial R_{k}}\frac{1}{R } &=&\frac{\partial (x_{k}-y_{k})}{\partial _{x_{k}}}\frac{\partial }{ \partial (x_{k}-y_{k})}\frac{\partial }{\partial R_{k}}\frac{1}{R}=\frac{ \partial }{\partial R_{k}}\frac{\partial }{\partial R_{k}}\frac{1}{R} \end{eqnarray*} Thus \begin{equation*} \partial _{\mathbf{x}}^{2}\frac{1}{R}=\partial _{\mathbf{R}}^{2}\frac{1}{R}% =-4\pi \delta ^{3}(\mathbf{R}) \end{equation*}