If $2\left(x-\frac{ydy}{dx}\right)\left(x^2+y^2\right)=\left(x^2-y^2\right)\left(y-\frac{xdy}{dx}\right),y\left(1\right)=0$ then

$2\left(x-\frac{ydy}{dx}\right)\left(x^2+y^2\right)=\left(x^2-y^2\right)\left(y-\frac{xdy}{dx}\right)$

$\Rightarrow \frac{2(xdx-ydy)}{x^2-y^2}=\frac{ydx-xdy}{x^2+y^2}$

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

$\Rightarrow \ln \left(x^2-y^2\right)+{\tan }^{-1}\frac{y}{x}=c,x=1,y=0\Rightarrow c=0$

$\Rightarrow \ln \left(x^2-y^2\right)+{\tan }^{-1}\frac{y}{x}=0$.