The curve satisfying the differential equation $\left(x^2-y^2\right)dx+2xydy=0$ and passing through the point ($1, 1$) is 

Given equation is  $\left(x^2-y^2\right)dx+2xydy=0\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}\to \left(1\right)$

Put $y=vx$. Then  $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\begin{array}{l}\text{ (1) }\Rightarrow v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2x^{2}v}=\frac{v^2-1}{2v}\Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=-\left(\frac{v^2+1}{2v}\right)\\ \Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}\Rightarrow \log \left(v^2+1\right)+\log x=\log c\Rightarrow x\left(v^2+1\right)=c\Rightarrow x\left(\frac{y^2+x^2}{x^2}\right)=c\\ \Rightarrow x^2+y^2=cx\end{array}$

$(1,1)\text{ lies on the curve }\Rightarrow 1+1=c\Rightarrow c=2$

$\therefore \text{ The curve is }{x}^2+y^2=2x$ which is a circle of radius one.