Answer
$${f_x}\left( {x,y} \right) = - \frac{{y{e^{ - xy}}}}{{1 + {e^{ - xy}}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{{x{e^{ - xy}}}}{{1 + {e^{ - xy}}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \ln \left( {1 + {e^{ - xy}}} \right) \cr
& {\text{Find the first partial derivative }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left( {1 + {e^{ - xy}}} \right)} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\frac{\partial }{{\partial x}}\left[ {1 + {e^{ - xy}}} \right] \cr
& {\text{use }}\left( {{e^u}} \right)' = {e^u}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\left( {{e^{ - xy}}} \right)\frac{\partial }{{\partial x}}\left[ { - xy} \right] \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\left( {{e^{ - xy}}} \right)\left( { - y} \right) \cr
& {f_x}\left( {x,y} \right) = - \frac{{y{e^{ - xy}}}}{{1 + {e^{ - xy}}}} \cr
& \cr
& {\text{Find the first partial derivative }}{f_x}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left( {1 + {e^{ - xy}}} \right)} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\frac{\partial }{{\partial y}}\left[ {1 + {e^{ - xy}}} \right] \cr
& {\text{use }}\left( {{e^u}} \right)' = {e^u}\left( {u'} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\left( {{e^{ - xy}}} \right)\frac{\partial }{{\partial y}}\left[ { - xy} \right] \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{1 + {e^{ - xy}}}}\left( {{e^{ - xy}}} \right)\left( { - x} \right) \cr
& {f_y}\left( {x,y} \right) = - \frac{{x{e^{ - xy}}}}{{1 + {e^{ - xy}}}} \cr} $$
