Answer
$$\frac{{dy}}{{dx}} = - \frac{{y{e^x}}}{{x{e^x} + y + 1}}e^y$$
Work Step by Step
$$\eqalign{
& {e^{xy}} + y{e^y} = 1 \cr
& {\text{Let }}f\left( {x,y} \right) = {e^{xy}} + y{e^y} \cr
& {\text{calculate the partial derivatives }}\frac{{\partial f}}{{\partial x}}{\text{ and }}\frac{{\partial f}}{{\partial y}} \cr
& \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{xy}} + y{e^y}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& \frac{{\partial f}}{{\partial x}} = y{e^{xy}} \cr
& and \cr
& \frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{e^{xy}} + y{e^y}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& \frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{e^{xy}}} \right] + y\frac{\partial }{{\partial y}}\left[ {{e^y}} \right] + {e^y}\frac{\partial }{{\partial y}}\left[ y \right] \cr
& \frac{{\partial f}}{{\partial y}} = x{e^{xy}} + y{e^y} + {e^y} \cr
& \cr
& {\text{Use the theorem 13}}{\text{.5}}{\text{.3 to find }}dy/dx \cr
& \frac{{dy}}{{dx}} = - \frac{{\partial f/\partial x}}{{\partial f/\partial y}} \cr
& \frac{{dy}}{{dx}} = - \frac{{y{e^{xy}}}}{{x{e^{xy}} + y{e^y} + {e^y}}} \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{{y{e^x}}}{{x{e^x} + y + 1}}e^y \cr} $$
