Answer
$-\dfrac{4}{3 \ln 2}$ and $-\dfrac{5}{3 \ln 2}$
Work Step by Step
Consider $F(x,y,z)=xe^y+ye^x+2 \ln x-2-3 \ln 2=0$
Now, $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$
$\implies \dfrac{\partial z}{ \partial x}=-(\dfrac{e^y+(2/x)}{ye^z})$
For the point $(1,\ln 2,\ln 3)$, we have :
$\dfrac{\partial z}{ \partial x}(1,\ln 2,\ln 3)=-\dfrac{e^{\ln 2}+\dfrac{2}{1}}{(\ln 2)e^{\ln 3}}=-\dfrac{4}{3 \ln 2}$
Now, consider $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$
$\implies \dfrac{\partial z}{ \partial y}=-\dfrac{xe^y+e^z}{ye^z}$
For the point $(1,\ln 2,\ln 3)$, we have :
$-[\dfrac{e^{\ln 2}+e^{\ln 3}}{(\ln 2)e^{\ln 3}}]=-\dfrac{5}{3 \ln 2}$
