Answer
$\ln xy=C+\frac{x^2}{2y^2}$
Work Step by Step
Given:
$$y(x^2-y^2)dx-x(x^2+y^2)dy=0$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$vx(x^2-v^2x^2)-x(x^2+v^2x^2)(xv'+v)=0$$
$$vx^3(1-v^2)-x^3(1+v^2)(xv'+v)=0$$
$$v(1-v^2)=(1+v^2)(xv'+v)$$
$$xv'=\frac{v(1-v^2)}{1+v^2}-v$$
$$\frac{v^2+1}{2v^3}dv+\frac{1}{x}dx=0$$
Integrating both sides:
$$\int \frac{v^2+1}{2v^3}dv=\frac{1}{2}(\int \frac{1}{v}dv+\int \frac{1}{v^3} dv$$
$$\int \frac{v^2+1}{2v^3}dv=\frac{\ln v}{2} - \frac{1}{4v^2}$$
so
$$\frac{\ln v}{2} - \frac{1}{4v^2}+\ln x=C$$
where $C$ is a constant of integration.
$$\frac{\ln vx^2}{2}=C+ \frac{1}{4v^2}$$
Subtitute when $y=vx \rightarrow \ln yx=2C+\frac{x^2}{2y^2}$
$$\ln xy=C+\frac{x^2}{2y^2}$$
