 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 identityThe 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.