Prove that $x^{2}-y^{2}=c(x^{2}+y^{2})^{2}$ is the general solution of the differential equation $(x^{3}-3xy^{2})dx=(y^{3}-3x^{2}y)dy,$ where $c$ is a parameter.

$(x^{3}-3xy^{2})dx=(y^{3}-3x^{2}y)dy$
$\Rightarrow \frac{dy}{dx}=\frac{x^{3}-3xy^{2}}{y^{3}-3x^{2}y}$ ...........$(1)$
This is a homogeneous equation. Let $y=vx$.
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting $y=vx$ in equation $(1)$:
$v+x\frac{dv}{dx}=\frac{x^{3}-3x(vx)^{2}}{(vx)^{3}-3x^{2}(vx)}=\frac{x^{3}-3xv^{2}x^{2}}{v^{3}x^{3}-3x^{3}v}=\frac{1-3v^{2}}{v^{3}-3v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-3v^{2}}{v^{3}-3v}-v=\frac{1-3v^{2}-v^{4}+3v^{2}}{v^{3}-3v}=\frac{1-v^{4}}{v^{3}-3v}$
$\Rightarrow \frac{v^{3}-3v}{1-v^{4}}dv=\frac{dx}{x}$
Integrating both sides:
$\int \frac{v^{3}-3v}{1-v^{4}}dv = \int \frac{dx}{x} + \log C'$
$\int \frac{v^{3}}{1-v^{4}}dv - 3\int \frac{v}{1-v^{4}}dv = \log x + \log C'$
$-\frac{1}{4}\log(1-v^{4}) - 3\int \frac{v}{1-(v^{2})^{2}}dv = \log(C'x)$
Let $v^{2}=p$,then $2vdv=dp$:
$-\frac{1}{4}\log(1-v^{4}) - \frac{3}{2}\int \frac{dp}{1-p^{2}} = \log(C'x)$
$-\frac{1}{4}\log(1-v^{4}) - \frac{3}{4}\log\left|\frac{1+v^{2}}{1-v^{2}}\right| = \log(C'x)$
Multiplying by $-4$:
$\log(1-v^{4}) + 3\log\left|\frac{1+v^{2}}{1-v^{2}}\right| = -4\log(C'x)$
$\log\left[(1-v^{2})(1+v^{2}) \cdot \frac{(1+v^{2})^{3}}{(1-v^{2})^{3}}\right] = \log(C'x)^{-4}$
$\frac{(1+v^{2})^{4}}{(1-v^{2})^{2}} = \frac{1}{C'^{4}x^{4}}$
Substituting $v=\frac{y}{x}$:
$\frac{(1+\frac{y^{2}}{x^{2}})^{4}}{(1-\frac{y^{2}}{x^{2}})^{2}} = \frac{1}{C'^{4}x^{4}} \Rightarrow \frac{(\frac{x^{2}+y^{2}}{x^{2}})^{4}}{(\frac{x^{2}-y^{2}}{x^{2}})^{2}} = \frac{1}{C'^{4}x^{4}}$
$\frac{(x^{2}+y^{2})^{4}}{x^{8}} \cdot \frac{x^{4}}{(x^{2}-y^{2})^{2}} = \frac{1}{C'^{4}x^{4}}$
$\frac{(x^{2}+y^{2})^{4}}{(x^{2}-y^{2})^{2}} = \frac{1}{C'^{4}} \Rightarrow (x^{2}-y^{2})^{2} = C'^{4}(x^{2}+y^{2})^{4}$
Taking square root:
$x^{2}-y^{2} = C(x^{2}+y^{2})^{2}$,where $C=C'^{2}$.