Answer
$\displaystyle \frac{\partial z}{\partial x}=\frac{\ln y}{2z-y},\qquad \displaystyle \frac{\partial z}{\partial y}=\frac{x+yz}{y(2z-y)}$
Work Step by Step
$yz+x\ln y=z^{2} \displaystyle \qquad/\frac{\partial}{\partial x}$, (y is constant, z is a function of x)
$\displaystyle \frac{\partial}{\partial x}(yz+x\ln y)=\frac{\partial}{\partial x}(z^{2})$
$y\displaystyle \frac{\partial z}{\partial x}+\ln y=2z\frac{\partial z}{\partial x}$
$\ln y =2z\displaystyle \frac{\partial z}{\partial x}-y\frac{\partial z}{\partial x}$
$\ln y= (2z -y)\displaystyle \frac{\partial z}{\partial x}$
$\displaystyle \frac{\partial z}{\partial x}=\frac{\ln y}{2z-y}$
$yz+x\ln y=z^{2} \displaystyle \qquad/\frac{\partial}{\partial y}$, (x is constant, z is a function of y)
$\displaystyle \frac{\partial}{\partial y}(yz+x\ln y)=\frac{\partial}{\partial y}(z^{2})$
$y\displaystyle \frac{\partial z}{\partial y}+z\cdot 1+x\cdot\frac{1}{y}=2z \displaystyle \frac{\partial z}{\partial y}$
$z +\displaystyle \frac{x}{y}=2z\frac{\partial z}{\partial y}-y\frac{\partial z}{\partial y}$
$\displaystyle \frac{yz+x}{y}=(2z-y) \displaystyle \frac{\partial z}{\partial y}$
$\displaystyle \frac{\partial z}{\partial y}=\frac{x+yz}{y(2z-y)}$.
