#### Answer

$$-\frac{ 2wz+3y}{x^2+3w^2+z^2}$$

#### Work Step by Step

Given $$ \quad x^{2} w+w^{3}+w z^{2}+3 y z=0$$
Consider $$F(x,y,z,w)=x^{2} w+w^{3}+w z^{2}+3 y z=0 $$
Then
\begin{align*}
F_w&= x^2+3w^2+z^2\\
F_z&= 2wz+3y
\end{align*}
Then
\begin{align*}
\frac{\partial w}{\partial z}&=-\frac{F_{z}}{F_{w}}\\
&=-\frac{ 2wz+3y}{x^2+3w^2+z^2}
\end{align*}