= Solution
Put $s=|\mathbf r-\mathbf r'|$. Away from the source, the radial <Laplacian> in three dimensions gives
$$
\Delta_{\mathbf r}G_0=\frac1{s^2}\frac d{ds}\left(s^2\frac d{ds}\left[-\frac1{4\pi s}\right]\right)
=\frac1{s^2}\frac d{ds}\frac1{4\pi}=0.
$$
Thus $\boxed{\Delta G_0=0\text{ for }\mathbf r\ne\mathbf r'}$. At the source it is a <distribution>, not a smooth function; its outward gradient flux through a small sphere is $4\pi s^2/(4\pi s^2)=1$, giving $\Delta G_0=\delta$.
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