Answer
$$y = {C_1}{e^{ - \frac{{{x^2}}}{{2{y^2}}}}},{\text{ }}{C_1} = \frac{1}{C}$$
Work Step by Step
$$\eqalign{
& xydx + \left( {{y^2} - {x^2}} \right)dy = 0 \cr
& {\text{Let }}y = ux,{\text{ }}dy = udx + xdu \cr
& {\text{Substituting}} \cr
& x\left( {ux} \right)dx + \left( {{u^2}{x^2} - {x^2}} \right)\left( {udx + xdu} \right) = 0 \cr
& u{x^2}dx + {u^3}{x^2}dx + {u^2}{x^3}du - u{x^2}dx - {x^3}du = 0 \cr
& {u^3}{x^2}dx + {u^2}{x^3}du - {x^3}du = 0 \cr
& {\text{Collect like terms}} \cr
& {u^3}{x^2}dx = {x^3}du - {u^2}{x^3}du \cr
& {u^3}{x^2}dx = \left( {1 - {u^2}} \right){x^3}du \cr
& {\text{Separate the variables}} \cr
& \frac{{{x^2}}}{{{x^3}}}dx = \frac{{1 - {u^2}}}{{{u^3}}}du \cr
& \frac{1}{x}dx = \left( {{u^{ - 3}} - \frac{1}{u}} \right)du \cr
& {\text{Integrate both sides}} \cr
& {\text{ln}}\left| x \right| + \ln \left| C \right| = \frac{{{u^{ - 2}}}}{{ - 2}} - \ln \left| u \right| \cr
& {\text{ln}}\left| {Cx} \right| = - \frac{1}{{2{u^2}}} - \ln \left| u \right| \cr
& {\text{ln}}\left| {Cx} \right| + \ln \left| u \right| = - \frac{1}{{2{u^2}}} \cr
& {\text{ln}}\left| {Cux} \right| = - \frac{1}{{2{u^2}}} \cr
& y = ux \to u = \frac{y}{x} \cr
& {\text{ln}}\left| {C\left( {\frac{y}{x}} \right)x} \right| = - \frac{{{x^2}}}{{2{y^2}}} \cr
& {\text{ln}}\left| {Cy} \right| = - \frac{{{x^2}}}{{2{y^2}}} \cr
& {\text{Solve for }}y \cr
& Cy = {e^{ - \frac{{{x^2}}}{{2{y^2}}}}} \cr
& y = {C_1}{e^{ - \frac{{{x^2}}}{{2{y^2}}}}},{\text{ }}{C_1} = \frac{1}{C} \cr} $$
