Answer
$\frac{\partial z}{\partial y}=\frac{-z - \frac{x}{y}}{y-2z}$
$\frac{\partial z}{\partial x}=\frac{\ln(y)}{y-2z}$
Work Step by Step
The function $yz+x\ln(y)=z^2$ is given. The partial derivatives of $z$ with respect to $x$ and $y$ need to be found.
Partial derivative w.r.t $x$:
$\frac{\partial}{\partial x} {(yz+x\ln(y))}= \frac{\partial}{\partial x} {z^2}$
$y \frac{\partial z}{\partial x} + \ln(y)=2z \frac{\partial z}{\partial x}$
Seperating the partial derivative:
$\frac{\partial z}{\partial x}=\frac{\ln(y)}{y-2z}$
Partial derivative w.r.t $y$:
$\frac{\partial}{\partial y} {(yz+x\ln(y))}= \frac{\partial}{\partial y} {z^2}$
$y \frac{\partial z}{\partial y} + z + \frac{x}{y}=2z \frac{\partial z}{\partial y}$
Seperating the partial derivative:
$\frac{\partial z}{\partial y}=\frac{-z - \frac{x}{y}}{y-2z}$
