The solution of $(x^2+xy)dy=(x^2+y^2)dx$ is:

$\log x=2\log (x-y)+\frac{y}{x}+c$

$\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}=\frac{1+(\frac{y}{x})^2}{1+\frac{y}{x}}$ Substitute y = xt $⇒\frac{dy}{dx}=t+x\frac{dt}{dx}⇒t+x,\frac{dt}{dx}=\frac{1+t^2}{1+t}$

$⇒x.\frac{dt}{dx}=\frac{1-t}{1+t}⇒\int\frac{1+t}{1-t}dt=\int\frac{dx}{x}⇒-\int\frac{1-t-2}{1-t}dt=\int\frac{dx}{x}$

$⇒-t+2\int\frac{dt}{1-t}=\int\frac{dx}{x}⇒-t-2ln(1-t)=ln\,x+c⇒\frac{-y}{x}-2ln(x-y)+ln\,x=c$