Answer
a) $\rho=\cot \phi \csc\phi$
b) $\rho=\cot \phi \csc\phi \sec 2 \theta $
Work Step by Step
The conversion of rectangular coordinates to spherical coordinates is given as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
a) Here, $z=x^2+y^2$
But $\rho=\sqrt {x^2+y^2+z^2}$
and $\rho \cos \phi=(\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2$
$\implies \rho \cos \phi=\rho^2 \sin^2 \phi$
$\rho=\cot \phi \csc\phi$
b) Here, we have $z=x^2-y^2$
But $\rho \cos \phi=(\rho \sin \phi \cos \theta)^2-(\rho \sin \phi \sin \theta)^2$
This gives: $\rho \cos \phi=\rho^2 \sin^2 \phi \cos 2 \theta$
and $\rho=\cot \phi \csc\phi \sec 2 \theta $
