Answer
$$\frac{{dy}}{{dx}} = - \frac{y}{x}$$
Work Step by Step
$$\eqalign{
& {e^{xy}} + xy = 1 \cr
& {\text{Differentiate each sides w}}{\text{.r}}{\text{.t}} \hspace{2mm} x \cr
& \frac{d}{{dx}}\left[ {{e^{xy}}} \right] + \frac{d}{{dx}}\left[ {xy} \right] = \frac{d}{{dx}}\left[ 1 \right] \cr
& {\text{Apply }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{ }} \cr
& {e^{xy}}\underbrace {\frac{d}{{dx}}\left[ {xy} \right]}_{{\text{Product rule}}} + \underbrace {\frac{d}{{dx}}\left[ {xy} \right]}_{{\text{Product rule}}} = \frac{d}{{dx}}\left[ 1 \right] \cr
& {e^{xy}}\left( {x\frac{{dy}}{{dx}} + y} \right) + x\frac{{dy}}{{dx}} + y = 0 \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& x{e^{xy}}\frac{{dy}}{{dx}} + y{e^{xy}} + x\frac{{dy}}{{dx}} + y = 0 \cr
& x{e^{xy}}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} = - y{e^{xy}} - y \cr
& \left( {x{e^{xy}} + x} \right)\frac{{dy}}{{dx}} = - y{e^{xy}} - y \cr
& \frac{{dy}}{{dx}} = \frac{{ - y{e^{xy}} - y}}{{x{e^{xy}} + x}} \cr
& {\text{Factor and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - y\left( {{e^{xy}} + 1} \right)}}{{x\left( {{e^{xy}} + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = - \frac{y}{x} \cr} $$
