Answer
The equations in spherical coordinates for this surface:
$\rho = 0$ or $\phi = \frac{\pi }{6}$ or $\phi = \frac{{5\pi }}{6}$.
Work Step by Step
We have ${z^2} = 3\left( {{x^2} + {y^2}} \right)$.
The relations between rectangular and spherical coordinates are
$x = \rho \sin \phi \cos \theta $
$y = \rho \sin \phi \sin \theta $
$z = \rho \cos \phi $
Substituting these in ${z^2} = 3\left( {{x^2} + {y^2}} \right)$ gives
${\left( {\rho \cos \phi } \right)^2} = 3\left( {{{\left( {\rho \sin \phi \cos \theta } \right)}^2} + {{\left( {\rho \sin \phi \sin \theta } \right)}^2}} \right)$
${\rho ^2}{\cos ^2}\phi = 3\left( {{\rho ^2}{{\sin }^2}\phi {{\cos }^2}\theta + {\rho ^2}{{\sin }^2}\phi {{\sin }^2}\theta } \right)$
${\rho ^2}{\cos ^2}\phi = 3{\rho ^2}{\sin ^2}\phi $
${\rho ^2}\left( {{{\cos }^2}\phi - 3{{\sin }^2}\phi } \right) = 0$
${\rho ^2} = 0$ ${\ \ }$ or ${\ \ }$ ${\cos ^2}\phi = 3{\sin ^2}\phi $, ${\ }$ $\tan \phi = \pm \frac{1}{{\sqrt 3 }}$
The solutions are the equations in spherical coordinates for this surface: $\rho = 0$ or $\phi = \frac{\pi }{6}$ or $\phi = \frac{{5\pi }}{6}$.
