Answer
\[
\begin{aligned}
\frac{2 x^{2}+y-z^{3}+3 z^{2}+3 w}{3}=\frac{\partial w}{\partial z} &
\end{aligned}
\]
\[
\begin{aligned}
-\frac{1}{3}=\frac{\partial w}{\partial y} &
\end{aligned}
\]
\[
\begin{aligned}
-\frac{4 x}{3}=\frac{\partial w}{\partial x} &
\end{aligned}
\]
Work Step by Step
We are given that
\[
z=\ln \left(3 w+ 2 x^{2}+y-z^{3}\right)
\]
Thus, w is a function of three variables, so differentiating it with respect to $x$ and taking y and z as constant:
\[
\begin{aligned}
\left(\ln \left(3 w+2 x^{2}+y-z^{3}\right)\right) \frac{\partial}{\partial x}&=\frac{\partial}{\partial x}(z) \\
0=\frac{1}{2 x^{2}+y-z^{3}+3 w} *\left(4 x+3 \frac{\partial w}{\partial x}\right) & \\
0=4 x+3 \frac{\partial w}{\partial x} & \\
&-\frac{4 x}{3}=\frac{\partial w}{\partial x}
\end{aligned}
\]
Differentiating it with respect to y:
\[
\begin{aligned}
\frac{\partial}{\partial y}(z) =\frac{\partial}{\partial y}\left(\ln \left(3 w+2 x^{2}+y-z^{3}\right)\right) &\\
0=\frac{1}{3 w+2 x^{2}+y-z^{3}} *\left(1+3 \frac{\partial w}{\partial y}\right) & \\
041+3 \frac{\partial w}{\partial y} & \\
-\frac{1}{3}=\frac{\partial w}{\partial y} &
\end{aligned}
\]
Differentiating it with respect to z:
\[
\begin{aligned}
\frac{\partial}{\partial z}(z)=\frac{\partial}{\partial z}\left(\ln \left(3 w+2 x^{2}+y-z^{3}\right)\right) & \\
1=\frac{1}{2 x^{2}+y-z^{3}+3 w} *\left(-3 z^{2}+3 \frac{\partial w}{\partial z}\right) & \\
-3 z^{2}+3 \frac{\partial w}{\partial z} &=3 w+2 x^{2}+y-z^{3} \\
\frac{2 x^{2}+y-z^{3}+3 z^{2}+3 w}{3}=\frac{\partial w}{\partial z} &
\end{aligned}
\]
