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