The surfaces given by the equation
$$(x^2+y^2)\cos^2 \psi + z^2 \cot^2 \psi = \alpha^2; 0<\psi<\pi$$ for $\psi$ constant are equipotential surfaces. I'm asked to express $\psi$ in terms of $x,y,z$. This seems basic yet I'm not seeing how to proceed. Can you provide guidance on the method/approach to use in solving problems of this type?
Define
$$r^{2}:=x^{2}+y^{2}$$
and
$$C:=cos(\psi )$$
using
$$cot^{2}(\psi ) =\frac{cos^{2}(\psi ) }{sin^{2}(\psi ) }=\frac{cos^{2}(\psi ) }{1-cos^{2}(\psi ) }$$
we get
$$r^{2}C^{2}+z^{2}\frac{C^{2}}{1-C^{2}}=\alpha ^{2}$$
which is equivalent to
$$r^{2}(1-C^{2})C^{2}+z^{2}C^{2}=(1-C^{2})\alpha ^{2}$$
which is just a quadratic in $C^{2}$. Note that $\cos(\psi )$ is invertible in the domain that you gave.
Can you take it from there?