Answer
$z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
Work Step by Step
We know that $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
Now, we will re-write the given equation as:
$yz+x \ln y-z^2=0$
Consider $F(x,y,z)=yz+x \ln y-z^2=0$
Now, $F_x=\ln y$ and $F_y=z+xy^{-1}$ and $F_z=y-2z$
$$z_x=-\dfrac{F_x}{F_z}=-\dfrac{\ln y}{y-2z} \\ z_y=-\dfrac{F_y}{F_z}=-\dfrac{z+xy^{-1}}{y-2z} $$
So, $z_x=\dfrac{\ln y}{2z-y}$ and $z_y=\dfrac{zy+x}{2yz-y^2}$
