Answer
$y^2=x^2\ln(\ln(Cx))$
Work Step by Step
Given:
$$2xydy-(x^2e^{-\frac{y^2}{x^2}}+2y^2)dx=0$$
$$\frac{dy}{dx}=\frac{x^2e^{-\frac{y^2}{x^2}}+2y^2}{2xy}$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x\frac{dv}{dx}+v=\frac{x^2e^{-\frac{y^2}{x^2}}+2(vx)^2}{2x(vx)}$$
$$x\frac{dv}{dx}+v=\frac{e^{-v^2}+2v^2}{2v}$$
$$x\frac{dv}{dx}=\frac{e^{-v^2}+2v^2}{2v}-v$$
$$x\frac{dv}{dx}=\frac{e^{-v^2}}{2v}$$
$$\frac{dx}{x}=\frac{2v}{e^{-v^2}}dv$$
Integrating both sides:
$$\int \frac{1}{x}dx=\int \frac{2v}{e^{-v^2}}dv$$
$$e^{v^2}=\ln x +C$$
where $C$ is a constant of integration.
Subtitute when $y=vx $
$$\rightarrow e^{\frac{y^2}{x^2}}=\ln x + C$$
$$\rightarrow y^2=x^2\ln(\ln x+C)$$
Let $C=\ln C$:
$$\rightarrow y^2=x^2\ln(\ln(Cx))$$
