Answer
$$\phi=\pi/3, \quad \phi=2\pi/3.$$
Work Step by Step
Since $y^2+x^2=3z^2$, and \begin{aligned}
& x=\rho \sin \phi \cos \theta\\
& y=\rho \sin \phi \sin \theta\\
& z=\rho \cos \phi,
\end{aligned} Then we have
$$ \rho^2\sin^2\phi(\sin^2\theta+\cos^2\theta)=3 \rho^2\cos^2\phi$$ and hence $$ \sin^2\phi=3 \cos^2\phi\Longrightarrow \tan\phi=\pm\sqrt{3}.$$ That is,$$\phi=\pi/3, \quad \phi=2\pi/3.$$
