Do $\frac{\partial{z}}{\partial{\phi}}=0$ implies $\frac{\partial{\phi}}{\partial{z}}=0$?

Question

If $\frac{dy}{dx}=0$, then $\frac{dx}{dy}$ is unbounded. I'v tried to derive $\nabla$ in sc and there is: $$ z=rcos(\theta)\\ \frac{\partial}{\partial{z}}=\frac{\partial{r}}{\partial{z}}\frac{\partial}{\partial{r}}+\frac{\partial{\theta}}{\partial{z}}\frac{\partial}{\partial{\theta}}+\frac{\partial{\phi}}{\partial{z}}\frac{\partial}{\partial{\phi}}\\ $$ Can I use $\frac{\partial{\phi}}{\partial{z}}=\frac{1}{\frac{\partial{z}}{\partial{\phi}}}$ here?