Prove That If $(x^{2}+y^{2})\cos^{2}\psi+z^{2}\cot^{2}\psi=A^2$ then $\nabla ^2 \psi=0$. Assume $$(x^{2}+y^{2})\cos^{2}\psi+z^{2}\cot^{2}\psi=A^2$$ which $A$ is constant. How we can show $\psi(x,y,z)$ satisfies the Laplacian equation $\psi_{xx}+\psi_{yy}+\psi_{zz}=0$ ($\operatorname{div}\nabla\psi=0$) without calculating $\psi(x,y,z)$? I calculate $\psi(x,y,z)$ itself and differentiate, but I'm looking for easier methods, It's not important to use what, only the time that it takes is important.
 A: I just want to introduce the proper implicit differentiation formula for this problem. If you don't want to go by directly finding $\psi$ then you should use implicit differentiation and I just give you the tools. For this purpose, you need a product rule and also a chain rule for the Laplacian operator and also a chain-rule for the gradient operator. First, define the following functions
$$\eqalign{
  & f = f(t)  \cr 
  & g = g(x,y,z)  \cr 
  & h = h(x,y,z) \cr} $$
Then the rules you need will be
$$\begin{align}
   {\nabla ^2}\left( {h  g} \right) &= g{\nabla ^2}h + h{\nabla ^2}g +         2\nabla h \cdot \nabla g  \cr 
    {\nabla ^2}\left( {f \circ g} \right) &= \left( {f'' \circ g} \right){\left\| {\nabla g} \right\|^2} + \left( {f' \circ g} \right){\nabla ^2}g  \cr 
   \nabla \left( {f \circ g} \right) &= \left( {f' \circ g} \right)\nabla g \cr\end{align}$$
You can prove them easily by using index notation or some vector identities. Now, combining the above relations by replacing $g$ with $f \circ g$ in the first formula leads to
$$\eqalign{
   {\nabla ^2}\left[ {h\left( {f \circ g} \right)} \right] &= \left( {f \circ g} \right){\nabla ^2}h + h{\nabla ^2}\left( {f \circ g} \right) + 2\nabla h \cdot \nabla \left( {f \circ g} \right)  \cr 
   &= \left( {f \circ g} \right){\nabla ^2}h + h\left[ {\left( {f'' \circ g} \right){{\left\| {\nabla g} \right\|}^2} + \left( {f' \circ g} \right){\nabla ^2}g} \right] + 2\nabla h \cdot \left[ {\left( {f' \circ g} \right)\nabla g} \right]  \cr } $$
and finally
$$\boxed{{\nabla ^2}\left[ {h\left( {f \circ g} \right)} \right] = \left( {f \circ g} \right){\nabla ^2}h + h\left( {f' \circ g} \right){\nabla ^2}g + 2\left( {f' \circ g} \right)\nabla h \cdot \nabla g + h\left( {f'' \circ g} \right){\left\| {\nabla g} \right\|^2}}$$
Then, just apply the Laplacian operator to the both side of
$$(x^{2}+y^{2})\cos^{2}\psi+z^{2}\cot^{2}\psi = A^{2}$$
to obtain 
$${\nabla ^2}\left[ {({x^2} + {y^2}){{\cos }^2}\psi } \right] + {\nabla ^2}\left[ {{z^2}{{\cot }^2}\psi } \right] = 0$$
and do the computations for proper choices of $h$, $f$, and $g$. The arithmetic seems to be lengthy so it takes some time! :)
