Answer
$$\left| x \right| = C{\left( {x - y} \right)^2}$$
Work Step by Step
$$\eqalign{
& \left( {x + y} \right)dx - 2xdy = 0 \cr
& {\text{Let }}y = ux,{\text{ }}dy = udx + xdu \cr
& {\text{Substituting}} \cr
& \left( {x + ux} \right)dx - 2x\left( {udx + xdu} \right) = 0 \cr
& {\text{Multiplying}} \cr
& xdx + uxdx - 2uxdx - 2{x^2}du = 0 \cr
& {\text{Collect like terms}} \cr
& \left( {xdx + uxdx - 2uxdx} \right) = 2{x^2}du \cr
& \left( {xdx - uxdx} \right) = 2{x^2}du \cr
& x\left( {1 - u} \right)dx = 2{x^2}du \cr
& {\text{Separate the variables}} \cr
& \frac{x}{{2{x^2}}}dx = \frac{2}{{1 - u}}du \cr
& \frac{1}{x}dx = \frac{2}{{1 - u}}du \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{x}} dx = 2\int {\frac{1}{{1 - u}}} du \cr
& \ln \left| x \right| + {C_1} = - 2\ln \left| {1 - u} \right| \cr
& {\text{Exponentiate both sides}} \cr
& {e^{\ln \left| x \right| + {C_1}}} = {e^{ - 2\ln \left| {1 - u} \right|}} \cr
& {e^{{C_1}}}\left| x \right| = {e^{ - 2\ln \left| {1 - u} \right|}} \cr
& y = ux \to u = \frac{y}{x} \cr
& C\left| x \right| = {e^{\ln {{\left| {1 - \frac{y}{x}} \right|}^{ - 2}}}} \cr
& C\left| x \right| = {\left( {\frac{{x - y}}{x}} \right)^{ - 2}} \cr
& C\left| x \right| = \frac{{{x^2}}}{{{{\left( {x - y} \right)}^2}}} \cr
& \left| x \right| = C{\left( {x - y} \right)^2} \cr} $$
