Laplacian identity. Consider $f$ and $g$ smooth functions. How to prove the following identity:
$$\Delta\left(\frac{f}{g}\right)=\frac{1}{g}\Delta f-\frac{2}{g}\nabla\left(\frac{f}{g}\right).\nabla g-\frac{f}{g^2}\Delta g, \ \ \ \ \ g\neq0$$
Thank you very much.
 A: I'm going to use $\nabla^2$ for the Laplacian, so herein, "$\nabla^2$" = "$\Delta$".
Start with the formula
$\nabla^2(uv) = v\nabla^2u + 2\nabla u \cdot \nabla v + u \nabla^2 v, \tag{0}$
a proof of which may be found in my answer to this question:  Show that $\nabla^2(fg)= f\nabla^2g+g\nabla^2f+2\nabla f\cdot\nabla g$.  Then, setting $f = u$ and $g^{-1} = v$, we have
$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f + 2\nabla f \cdot \nabla (g^{-1}) + f\nabla^2(g^{-1}); \tag{1}$
we proceed by first calculating $\nabla (g^{-1})$ and then $\nabla^2(g^{-1})$.  $\nabla (g^{-1})$ looks like the easier one; indeed, for any differentiable function $G(g)$,
$\nabla (G(g)) = G'(g)\nabla g, \tag{2}$
which easily follows from the chain rule; for example, in Cartesian coordinates $x_i$ on $\Bbb R^n$, with 
$\nabla w = (\dfrac{\partial w}{\partial x_1}, \dfrac{\partial w}{\partial x_2}, . . . , \dfrac{\partial w}{\partial x_n})^T; \tag{3}$
taking $w = G(g)$, we have
$\dfrac{\partial G(g)}{\partial x_i} = \dfrac{d G}{d g}\dfrac{\partial g}{\partial x_i} = G'(g)\dfrac{\partial g}{\partial x_i}, \tag{4}$
from which (2) readily follows.  Equation (2) may also be established via the standard, coordinate-independent definition of $\nabla w$ as the unique vector field satisfying
$\langle \nabla w, X \rangle = X[w] \tag{5}$
for all vector fields $X$, where $\langle \cdot, \cdot \rangle$ is the inner product; for then
$\langle \nabla G(g), X \rangle = X[G(g)] = G'(g)X[g] = G'(g)\langle \nabla g, X \rangle = \langle G'(g) \nabla g , X \rangle, \tag{6}$
demonstrating (2) by a somewhat different argument.  Taking $G(g) = g^{-1}$ then yields
$\nabla (g^{-1}) = - g^{-2}\nabla g. \tag{7}$
We move on to finding $\nabla^2 (g^{-1})$:
$\nabla^2 (g^{-1}) = \nabla \cdot \nabla  (g^{-1}) = \nabla \cdot (-g^{-2} \nabla g), \tag{8}$
using (7); we evaluate $\nabla \cdot (-g^{-2} \nabla g)$ by exploiting the formula
$\nabla \cdot (wX) = \nabla w \cdot X + w \nabla \cdot X \tag{9}$
which may also be found in the linked question cited above, along with some explanatory remarks.  (8) and (9) yield, with the aid of (2),
$\nabla^2 (g^{-1}) = \nabla \cdot (-g^{-2} \nabla g) = \nabla (-g^{-2}) \cdot \nabla g -g^{-2} \nabla^2g = 2g^{-3}\nabla g \cdot \nabla g -g^{-2} \nabla^2g, \tag{10}$
and when (7) and (10) are inserted into (1) we find that
$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f + 2\nabla f \cdot (- g^{-2}\nabla g) + f(2g^{-3}\nabla g \cdot \nabla g -g^{-2} \nabla^2g)$
$= g^{-1}\nabla^2 f - fg^{-2} \nabla^2g - 2g^{-1}(\nabla f \cdot g \nabla g - fg^{-2}\nabla g \cdot \nabla g)$
$= g^{-1}\nabla^2 f - fg^{-2} \nabla^2g - 2g^{-1}(g \nabla f - fg^{-2}\nabla g)\cdot \nabla g. \tag{11}$
We finally evaluate the expression $g \nabla f - fg^{-2}\nabla g$ which occurs inside the parentheses in the last line of (11), using the formula
$\nabla (uv) = u\nabla v + v \nabla u, \tag{12}$
a variant of the ordinary Leibniz rule for the derivative of a product, a discussion of which is also available in the linked citing.  From (12) and yet again (2),
$\nabla (fg^{-1}) = g^{-1}\nabla f + f\nabla g^{-1} = g^{-1}\nabla f - fg^{-2}\nabla g. \tag{13}$
Using (13) in (11) and performing some minor re-arrangements we finally arrive at our destination, or perhaps I should say, like Darth Vader in Star Wars, our Destiny:
$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f - 2g^{-1}\nabla (fg^{-1}) \cdot \nabla g - fg^{-2}\nabla^2 g. \tag{14}$
(14) is, of course, the requisite result.
Hope this helps.  May The Force be with us, one and all,
and of course,
Fiat Lux!!!
A: Apply the product rule 
$$\Delta(uv)=u\Delta v+2\nabla u\cdot\nabla v+v\Delta u \tag{1}$$
with $u=f/g$ and $v=g$. The result is 
$$\Delta f=\frac{f}{g}\Delta g+2\nabla \left(\frac{f}{g}\right)\cdot \nabla g+g\,\Delta \left(\frac{f}{g}\right)$$
which can be rearranged as
$$\Delta \left(\frac{f}{g}\right)
=\frac{1}{g}\Delta f-\frac{2}{g}\nabla \left(\frac{f}{g}\right)\cdot \nabla g-\frac{f}{g^2}\Delta g$$
