Answer
$$
yz+x\ln y=z^{2}
$$
To find $\frac{\partial z}{\partial x}$, we differentiate implicitly with respect to $x$, being careful to treat $y$ as a constant:
$$
\begin{aligned}
\frac{\partial z}{\partial x} &= \frac{\ln y}{(2 z -y )} \\
\end{aligned}
$$
To find $\frac{\partial z}{\partial y}$, we differentiate implicitly with respect to $y$, being careful to treat $x$ as a constant:
$$
\begin{aligned}
\frac{\partial z}{\partial y} &=\frac{x+y z}{y(2 z-y)}
\end{aligned}
$$
Work Step by Step
$$
yz+x\ln y=z^{2}
$$
To find $\frac{\partial z}{\partial x}$, we differentiate implicitly with respect to $x$, being careful to treat $y$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial x} (yz+x\ln y) &=\frac{\partial }{\partial x} (z^{2})\\
y \frac{\partial z}{\partial x}+\ln y &=2 z \frac{\partial z}{\partial x}
\\
\ln y &=2 z \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial x} \\
\ln y &=(2 z -y )\frac{\partial z}{\partial x} \\
\\
\frac{\partial z}{\partial x} &= \frac{\ln y}{(2 z -y )} \\
\end{aligned}
$$
To find $\frac{\partial z}{\partial y}$, we differentiate implicitly with respect to $y$, being careful to treat $x$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial y} (yz+x\ln y) &=\frac{\partial }{\partial y} (z^{2}) \\
y \frac{\partial z}{\partial y}+z.1+x\frac{1}{y} &=2 z \frac{\partial z}{\partial x}
\\
z+\frac{x}{y} &=2 z \frac{\partial z}{\partial x} -y \frac{\partial z}{\partial y}
\\
z+\frac{x}{y} &=(2 z -y) \frac{\partial z}{\partial y}\\
\frac{\partial z}{\partial y} &=\frac{z+(x / y)}{2 z-y} \\
&=\frac{x+y z}{y(2 z-y)}
\end{aligned}
$$
