Answer
$r = \sqrt{x^2 + y^2 + z^2}$
$\theta = cos^{-1}(\frac{z}{\sqrt{x^2 + y^2 + z^2}})$
$\phi = tan^{-1}(\frac{y}{x})$
Work Step by Step
Beginning with the following :
$x = rsin\theta cos\phi \ \ y = r sin\theta sin\phi \ \ z = r cos\theta$
We can derive $r,\theta,\phi$ as follows
$x^2 + y^2 + z^2 = r^2(sin^2\theta cos^2\phi + sin^2\theta sin^2\phi + cos^2\theta)$
$ = r^2(sin^2\theta)(cos^2\phi + sin^2\phi)+ cos^2\theta$
$ = r^2(sin^2\theta + cos^2\theta) = r^2$
$r = \sqrt{x^2 + y^2 + z^2}$
From $z = r cos\theta$ we can substitute the above expression for r and solve for $\theta$
$\theta = cos^{-1}({\frac{z}{\sqrt{x^2 + y^2 + z^2}}})$
If we reduce the sphere to a circle on the xy plane (where $\phi$ is defined) and enforce $z = 0, \theta = \frac{\pi}{2}$ we have the familiar polar equations
$x = rcos\phi \ \ y = rsin\phi \ \ r^2 = x^2 + y^2$
$\frac{y}{x} = \frac{rsin\phi}{rcos\phi} = tan\phi$, so:
$\phi = tan^{-1}(\frac{y}{x})$
