Solve $\left(x^2-y^2\right)dx-xydy=0$.

$\left(x^2-y^2\right)dx-xydy=0$

$\frac{dy}{dx}=\frac{x^2-y^2}{xy}$.....(1)

It is a homogeneous equation to solve this equation

Put $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$...(2)

From (1) and (2)

$v+x\frac{dv}{dx}=\frac{x^2-v^2{x}^2}{v{x}^2}\Rightarrow v+x\frac{dv}{dx}=\frac{1-v^2}{v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{v}-\frac{v}{1}=\frac{1-v^2-v^2}{v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1-2v^2}{v}$

Separating the variables and integrating

$\int \frac{vdv}{1-2v^2}=\int \frac{dx}{x}$

Put  $1-2v^2=t\Rightarrow -4vdv=dt\Rightarrow vdv=\frac{-1}{4}dt$

$\therefore -\frac{1}{4}\int \frac{dt}{t}=\int \frac{dx}{x}\Rightarrow -\frac{1}{4}\log \left|t\right|=\log x+\log c$

$\Rightarrow -\frac{1}{4}\log \left|1-2v^2\right|=\log x+\log c$

$\Rightarrow -\frac{1}{4}\log \left|1-\frac{2y^2}{x^2}\right|=\log x+\log c$

$\Rightarrow -\frac{1}{4}\left[\log \left(x^2-2y^2\right)-\log x^2\right]=\log x+\log c$

$\Rightarrow -\frac{1}{4}\log \left(x^2-2y^2\right)+\frac{1}{4}\cdot 2\log x=\log x+\log c$

$\Rightarrow -\frac{1}{4}\log \left(x^2-2y^2\right)=\frac{1}{2}\log x+\log c$

$\Rightarrow \log \left(x^2-2y^2\right)=-2\log x-4\log c$

$\Rightarrow \log \left(x^2-2y^2\right)=-2\log x+\log k$

$\Rightarrow \log \left(x^2-2y^2\right)+\log x^2=\log k$

$\Rightarrow \log k=\log x^2\left(x^2-2y^2\right)$

$\Rightarrow k=x^2\left(x^2-2y^2\right)$