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