Answer
$z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
Work Step by Step
We are given: $yz+x \ln y=z^2$
Re-arrange as: $yz+x \ln y-z^2=0$
Consider $F(x,y,z)=yz+x \ln y-z^2=0$
$F_x=\ln y$ and $F_y=z+xy^{-1}$ and $F_z=y-2z$
Use Equation 7: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
$z_x=-\dfrac{F_x}{F_z}=-\dfrac{\ln y}{y-2z}$ and $z_y=-\dfrac{F_y}{F_z}=-\dfrac{z+xy^{-1}}{y-2z}$
Hence, we have $z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
