The solution of  $\left(x^2+y^2\right)dx=2xydy$ is

Put  $y=vx.\text{ Then }\frac{dy}{dx}=v+x\frac{dv}{dx}.\Rightarrow \left(x^2+y^2\right)dx=2xydy\Rightarrow \frac{dy}{dx}=\frac{x^2+y^2}{2xy}$

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