= Solution
For $s=|\mathbf x-\mathbf x_0|>0$, the radial <Laplacian> gives $\nabla^2(1/s)=s^{-2}\partial_s(s^2\partial_s(1/s))=0$. The outward <flux> of $\nabla[-1/(4\pi s)]$ through a small <sphere> centred at the source is $[1/(4\pi s^2)]4\pi s^2=1$. Applying <Green's second identity> to a smooth <test function> outside that small sphere and shrinking its radius therefore gives the distributional identity
$$
\boxed{\nabla^2\left(-\frac1{4\pi|\mathbf x-\mathbf x_0|}\right)=\delta(\mathbf x-\mathbf x_0).}
$$
The kernel tends to zero at infinity. Thus it is the <fundamental solution> with the sign convention $\Delta G=+\delta$. The sign will reverse when converting it into the <electrostatic potential> of a positive <point charge> through <Poisson's equation>.
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