Calculating a term using a multilinear map How to calculate the value of the term
$\Delta u:=u_{xx}+u_{yy}+u_{zz}=\large\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}$
for the function $u$ on $\mathbb{R}^3 \backslash \{0\}$ with
$u(x,y,z):=\large\frac{1}{\sqrt{x^2+y^2+z^2}}$  ?
 A: If you know the laplacian in spherical coordinates
$$\Delta u= {1 \over r^2} {\partial \over \partial r}
  \left(r^2 {\partial u \over \partial r} \right) 
+ {1 \over r^2 \sin \theta} {\partial \over \partial \theta}
  \left(\sin \theta {\partial u \over \partial \theta} \right) 
+ {1 \over r^2 \sin^2 \theta} {\partial^2 u \over \partial \varphi^2}, $$
since $u$ in your case is the radial function $1/r$, all you have to do is find
$$\frac{1}{r^2}(r^2 u_r)_r=\frac{1}{r^2}(-1)_r=0. $$ 
A: Answer

$\nabla^2{\left(\frac{1}{\sqrt{x^2 + y^2 + z^2}} \right)} = -4\,\pi\,\delta{\left(\mathbf{x}\right)}$

Context
The Laplacian of OP's $u$ has a singular nature. This singular nature is not captured by @user1337's answer.  As explained in [1], we can use a limiting process to integrate around the singularity as follows.
Derivation
By $B_r{(\mathbf{0})} \subset \mathbb{R}^3$, I denote an open ball of radius $r > 0$ centered at a point $\mathbf{0}$. By $S^2{(r)}$, I denote a 2-sphere of radius $r$. Then,
\begin{align}
\int_{B_r{(\mathbf{0})}} \nabla^2{u} \left|d\mathbf{x}^3\right|
&=
\int_{S^2{(r)}} \mathbf{n}\cdot  \boldsymbol{\nabla}{u}  \, dS 
&&
\text{divergence theorem}
\\
&=
\int_{S^2{(r)}} \mathbf{n}\cdot\boldsymbol{\nabla}{\frac{1}{\sqrt{x^2 + y^2 + z^2}}}  \, dS 
&&
\text{substituting given}
\\
&=
\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot\boldsymbol{\nabla}{\frac{1}{r}}  \, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{radial coordinates}
\\
&=
\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot
\frac{\partial \frac{1}{r}}{\partial r}   \mathbf{\widehat{r}}
\, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{gradient }
\\
&=
-\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot
  \frac{1}{r^2}    \mathbf{\widehat{r}}
\, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{derivative }
\\
&=
-\int_{S^2{(r)}}  
   \sin{\theta}\, d\theta\,d\phi 
&&
\text{simplifying }
\\
&=
-4\,\pi
\end{align}
From [1], ``It is now established that $\nabla^2 {\left(1/r\right)}$ is 0 for $r\neq 0$, and that its volume is $-4\,\pi$. Consequently, we can write the improper [but mathematically justifiable] equation $\nabla^2{(1/r)} = -4\,\pi\,\delta{\left(\mathbf{x}\right)}$''
Bibliography
[1] Jackson, John D. Classical Electrodynamics
