Answer
$z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
Work Step by Step
Recall Equation 7: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ ...(1)
Given: $yz+x \ln y=z^2$
This gives: $yz+x \ln y-z^2=0$
Let us consider: $F(x,y,z)=yz+x \ln y-z^2=0$
$F_x=\ln y\\F_y=z+xy^{-1}\\F_z=y-2z$
Equation (1) becomes: $z_x=-\dfrac{\ln y}{y-2z}$ and $z_y=-\dfrac{z+xy^{-1}}{y-2z}$
Therefore, $z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
