Answer
$\frac{\partial z}{\partial x}=-\frac{y^{2}(1+z)}{(2+z)}$
$\frac{\partial z}{\partial y}=-\frac{2 x y(1+z)}{(2+z)}$
Work Step by Step
\[
1=\ln (z+1)+x y^{2}+z
\]
We have to find $\frac{\partial z}{\partial y}$ and $\frac{\partial z}{\partial x}$
\[
\begin{array}{l}
\frac{\partial z}{\partial x}=-\frac{\partial f / \partial x}{\partial f / \partial z} \\
\frac{\partial z}{\partial y}=-\frac{\partial f / \partial y}{\partial f / \partial z}
\end{array}
\]
We use this theorem in order to find it
\[
y^{2}=\frac{\partial f}{\partial x}
\]
\[
\begin{array}{l}
\frac{\partial f}{\partial y}=2 x y \\
\frac{\partial f}{\partial z}=\frac{z+2}{z+1}
\end{array}
\]
Substituting:
$\frac{\partial z}{\partial x}=-\frac{y^{2}(1+z)}{(2+z)}$
$\frac{\partial z}{\partial y}=-\frac{2 x y(1+z)}{(2+z)}$
