Answer
\[
\begin{array}{l}
\frac{1}{3 z^{2}} =\frac{\partial z}{\partial y}\\
\frac{-2 x^{2}-y+z^{3}+4 x}{3 z^{2}} =\frac{\partial z}{\partial x}
\end{array}
\]
Work Step by Step
We are given that
\[
X=\ln \left(-z^{3}+y+2 x^{2}\right)
\]
Differentiate the variable $ z $ in relation to $ x $ with $ y $ constant
\[
\begin{aligned}
\frac{\partial}{\partial x}\left(\ln \left(2 x^{2}+y-z^{3}\right)\right) &=\frac{d}{d x}(x) \\
1=\frac{1}{-z^{3}+y+2 x^{2}} *\left(-3 z^{2} +4 x \frac{\partial z}{\partial x}\right) & \\
&-z^{3}+y+ 2 x^{2}=4 x-3 z^{2} \frac{\partial z}{\partial x} \\
&\frac{-2 x^{2}-y+z^{3}+4 x}{3 z^{2}}=\frac{\partial z}{\partial x}
\end{aligned}
\]
Similarly, distinguishing between the variable variable $ z $ in relation to $ y $ with $ x $ as a constant
\[
\begin{aligned}
\frac{\partial}{\partial y}\left(\ln \left(2 x^{2}+y-z^{3}\right)\right) &=\frac{d}{d y}(x) \\
\frac{1}{2 x^{2}+y-z^{3}} *\left(1-3 z^{2} \frac{\partial z}{\partial y}\right) &=0 \\
1-3 z^{2} \frac{\partial z}{\partial y} &=0 \\
\frac{1}{3 z^{2}}=\frac{\partial z}{\partial y} &
\end{aligned}
\]
