It is often quoted in physics textbooks for finding the electric potential using Green's function that
$$\nabla ^2 \left(\frac{1}{r}\right)=-4\pi\delta^3({\bf r}),$$
or more generally
$$\nabla ^2 \left(\frac{1}{|| \vec x - \vec x'||}\right)=-4\pi\delta^3(\vec x - \vec x'),$$
where $\delta^3$ is the 3-dimensional Dirac delta distribution. However I don't understand how/where this comes from. Would anyone mind explaining?
The gradient of $\frac1r$ (noting that $r=\sqrt{x^2+y^2+z^2}$) is
$$ \nabla \frac1r = -\frac{\mathbf{r}}{r^3} $$ when $r\neq 0$, where $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$. Now, the divergence of this is
$$ \nabla\cdot \left(-\frac{\mathbf{r}}{r^3}\right) = 0 $$ when $r\neq 0$. Therefore, for all points for which $r\neq 0$,
$$ \nabla^2\frac1r = 0 $$ However, if we integrate this function over a sphere, $S$, of radius $a$, then, applying Gauss's Theorem, we get
$$ \iiint_S \nabla^2\frac1rdV = \iint_{\Delta S} -\frac{\mathbf{r}}{r^3}.d\mathbf{S} $$ where $\Delta S$ is the surface of the sphere, and is outward-facing. Now, $d\mathbf{S}=\mathbf{\hat r}dA$, where $dA=r^2\sin\theta d\phi d\theta$. Therefore, we may write our surface integral as $$\begin{align} \iint_{\Delta S} -\frac{\mathbf{r}}{r^3}.d\mathbf{S}&=-\int_0^\pi\int_0^{2\pi}\frac{r}{r^3}r^2\sin\theta d\phi d\theta\\ &=-\int_0^\pi\sin\theta d\theta\int_0^{2\pi}d\phi\\ &= -2\cdot 2\pi = -4\pi \end{align}$$ Therefore, the value of the laplacian is zero everywhere except zero, and the integral over any volume containing the origin is equal to $-4\pi$. Therefore, the laplacian is equal to $-4\pi \delta(\mathbf{r})$.
EDIT: The general case is then obtained by replacing $r=|\mathbf{r}|$ with $s=|\mathbf{r}-\mathbf{r_0}|$, in which case the function shifts to $-4\pi \delta(\mathbf{r}-\mathbf{r_0})$