Given
$$\vec{\nabla}\cdot\vec{E}=\frac{1}{r}\frac{\partial}{\partial r}\left[r\left(\frac{az}{r}\right)\right]\frac{1}{r}\frac{\partial}{\partial\phi}(br)+\frac{\partial}{\partial z}(cr^2z^2)=\frac{\rho}{\varepsilon_0},$$
why are there two zeros in this resulting equation? $$\frac{1}{r}(0)+\frac{1}{r}(0)+2zcr^2=\frac{\rho}{\varepsilon_0}$$
The functions inside of the derivative are constant with respect to $r$ and $\phi$, and thus have a zero derivative.
Note that in the case of the left partial derivative the $r$'s cancel as below, thus yielding the result.
$$\frac{\partial}{\partial r} \left[ r \left( \frac{az}{r} \right) \right] = \frac{\partial}{\partial r} \left[az \right] = 0$$