Answer
$\rightarrow (x-y)=C(y-2x)^2(x+y)$
Work Step by Step
Given:
$$\frac{dy}{dx}=\frac{y^2+2xy-2x^2}{x^2-xy+y^2}$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x\frac{dv}{dx}+v=\frac{(vx)^2+2x(vx)-2x^2}{x^2-x(vx)+(vx)^2}$$
$$x\frac{dv}{dx}+v=\frac{v^2+2v-2}{v^2-v+1}$$
$$x\frac{dv}{dx}=\frac{v^2+2v-2}{v^2-v+1}-v$$
$$x\frac{dv}{dx}=\frac{v^3-2v^2-v+2}{-v^2+v-1}$$
$$\frac{dx}{x}=\frac{-v^2+v-1}{v^3-2v^2-v+2}dv$$
Integrating left sides:
$$\int\frac{-v^2+v-1}{v^3-2v^2-v+2}dv=\int (\frac{1}{2(v-1)}-\frac{1}{2(v+1)}-\frac{1}{v-2}dv=\frac{1}{2}\ln (v-1)-\frac{1}{2}\ln (v+1)-\ln (v-2)=\frac{1}{2}\ln [\frac{v-1}{(v+1)(v-2)^2}]$$
Integrating both sides, hence:
$$\frac{1}{2}\ln [\frac{v-1}{(v+1)(v-2)^2}]=\ln x +ln C$$
where $C$ is a constant of integration.
$$\ln [\frac{v-1}{(v+1)(v-2)^2}]=2\ln x+ \ln C$$
$$\rightarrow [\frac{v-1}{(v+1)(v-2)^2}]=Cx^2$$
Subtitute when $y=vx $
$$\rightarrow \frac{1-\frac{y}{x}}{(\frac{y}{x}-2)^2(\frac{y}{x}+1)}=Cx^2$$
$$\rightarrow \frac{x^2(x-y)}{(y-2x)^2(x+y)}=Cx^2$$
$$\rightarrow (x-y)=C(y-2x)^2(x+y)$$
