Let  $f_1 \equiv x - u - e^{-v} \text{ and } f_2 \equiv y - v - e^{-u}$

$\dfrac {\partial u}{\partial y} = - \left [ \dfrac {\frac {\partial (f_1, f_2)}{\partial (y, v)}}{\frac {\partial (f_1, f_2)}{\partial (u, v)}} \right ]$ and $ \dfrac {\partial v}{\partial x} = - \left [ \dfrac {\frac {\partial (f_1, f_2)}{\partial (u, x)}} { \frac {\partial (f_1, f_2)}{(\partial (u, v))}}\right ]$

Now,

$\dfrac {\partial (f_1, f_2)}{\partial (y, v)} = \begin{vmatrix} 0 &e^{-v} \\1 &-1 \end{vmatrix} = e^{-v}$

$\dfrac {\partial (f_1 , f_2)}{\partial (u, v)} = \begin{vmatrix} -1 &e^{-v} \\e^{-u} &-1 \end{vmatrix} = e^{u/v}-1$

$ And \dfrac {\partial (f_1, f_2)}{\partial (u, x)} = \begin{vmatrix} -1 &1 \\e^{-u} &0 \end{vmatrix} = e^{-u}$

Hence,

$\dfrac {\partial u}{\partial y} = - \left [ \dfrac {e^{-v}}{e^{uv}-1} \right ]$

$\dfrac {\partial v}{\partial x} = - \left [ \dfrac {e^{-u}}{e^{uv}-1} \right ]$