Answer
Please see the figure attached.
Work Step by Step
We have $\rho = \csc \phi \cot \phi $. Write
$\rho = \frac{1}{{\sin \phi }}\frac{{\cos \phi }}{{\sin \phi }}$
(1) ${\ \ \ }$ $\rho \sin \phi = \frac{{\cos \phi }}{{\sin \phi }}$
Recall the relations between cylindrical and spherical coordinates can be found using rectangular coordinates $x$, $y$, $z$ as in the following:
$x = r\cos \theta = \rho \sin \phi \cos \theta $
$y = r\sin \theta = \rho \sin \phi \sin \theta $
$z = \rho \cos \phi $
So, $r = \rho \sin \phi $ and $z = \rho \cos \phi $. Whereas $\theta$ is the same in both cylindrical and spherical coordinates.
Dividing $z$ by $r$ gives
$\frac{z}{r} = \frac{{\cos \phi }}{{\sin \phi }}$
Thus, equation (1) becomes
$r = \frac{z}{r}$, ${\ \ \ }$ $z = {r^2}$
Since ${r^2} = {x^2} + {y^2}$,
$z = {x^2} + {y^2}$
This is the equation of an elliptic paraboloid (see Eq. (3) of Section 13.6).
