Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& x - y = xy \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ x \right] - \frac{d}{{dx}}\left[ y \right] = \underbrace {\frac{d}{{dx}}\left[ {xy} \right]}_{{\text{Product rule}}} \cr
& \frac{d}{{dx}}\left[ x \right] - \frac{d}{{dx}}\left[ y \right] = x\frac{d}{{dx}}\left[ y \right] + y\frac{d}{{dx}}\left[ x \right] \cr
& 1 - \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y\left( 1 \right) \cr
& 1 - \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} = 1 - y \cr
& \left( {1 + x} \right)\frac{{dy}}{{dx}} = 1 - y \cr
& \frac{{dy}}{{dx}} = \frac{{1 - y}}{{1 + x}} \cr
& \cr
& {\text{Solving for }}y \cr
& x - y = xy \cr
& xy + y = x \cr
& y\left( {x + 1} \right) = x \cr
& y = \frac{x}{{x + 1}} \cr
& {\text{Differentiate by using the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr
& \cr
& {\text{Substituting }}y = \frac{x}{{x + 1}}{\text{ into }}\frac{{dy}}{{dx}} = \frac{{1 - y}}{{1 + x}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 - \frac{x}{{x + 1}}}}{{1 + x}} \cr
& \frac{{dy}}{{dx}} = \frac{{x + 1 - x}}{{{{\left( {1 + x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}} \cr} $$
