Answer
$-\dfrac{4}{3 \ln 2}$ and $-\dfrac{5}{3 \ln 2}$
Work Step by Step
Since, we have $F(x,y,z)=xe^y+ye^x+2 \ln x-2-3 \ln 2=0$
$\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{\partial z}{ \partial x}=-\dfrac{e^y+\dfrac{2}{x}}{ye^z}$
Now, plug the point $(1,\ln 2,\ln 3)$
Thus, we get $\dfrac{\partial z}{ \partial x}=-\dfrac{e^{\ln 2}+\dfrac{2}{1}}{(\ln 2)e^{\ln 3}}=-\dfrac{4}{3 \ln 2}$
Now, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ ...(2)
Plug the derivatives in equation (2).
we have $\dfrac{\partial z}{ \partial y}=-\dfrac{xe^y+e^z}{ye^z}$
and $\dfrac{\partial z}{ \partial y}(1,\ln 2,\ln 3)=-\dfrac{e^{\ln 2}+e^{\ln 3}}{(\ln 2)e^{\ln 3}}=-\dfrac{5}{3 \ln 2}$
