The Laplace operator. What will be the value of $\Delta\left(\frac 1{r^2}\right)$ if $r=|x|=\sqrt{x_1^2+x_2^2+x_3^2}$?
May be this can be determined using Green's formula.

If we don't wanna use coordinate transform, it can be also done in a "hands-on" way:
$$\Delta u= \sum_{i}\frac{\partial^2 u}{\partial x_i^2},$$
where $$u = \frac{1}{r^2} = \frac{1}{\sum\limits_i x_i^2}= \frac{1}{x_1^2 + x_2^2 + x_3^2}.$$
We can compute it term by term:
$$
\frac{\partial u}{\partial x_1} = -\frac{1}{(\sum_i x_i^2)^2}\cdot 2 x_1 = -\frac{ 2 x_1}{(x_1^2 + x_2^2 + x_3^2)^2} ,
$$
and
$$
\frac{\partial^2 u}{\partial x_1^2} = -\frac{2 (\sum_i x_i^2) ^2 - 2 x_1 \cdot2(\sum_i x_i^2) 2 x_1 }{(\sum_i x_i^2)^4} =  -\frac{2 (x_1^2 + x_2^2 + x_3^2)^2- 8 x_1^2 (x_1^2 + x_2^2 + x_3^2) }{(x_1^2 + x_2^2 + x_3^2)^4} ,
$$
Hence:
$$
\Delta u = -\frac{6 (x_1^2 + x_2^2 + x_3^2)^2- 8  (x_1^2 + x_2^2 + x_3^2)^2 }{(x_1^2 + x_2^2 + x_3^2)^4} = \frac{2}{(x_1^2 + x_2^2 + x_3^2)^2} = \frac{2}{r^4}.
$$
