Answer
$\rightarrow y^2x=Ce^{\frac{4x}{y}}$
Work Step by Step
Given:
$$2x(y+2x)y'=y(4x-y)$$
$$\frac{dy}{dx}=\frac{y(4x-y)}{2x(y+2x)}$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x\frac{dv}{dx}+v=\frac{(4-v)v}{2(v+2)}$$
$$x\frac{dv}{dx}=\frac{(4-v)v}{2(v+2)}-v$$
$$x\frac{dv}{dx}=\frac{-3v^2}{2v+4}$$
$$\frac{-2(v+2)}{3v^2}dv=\frac{1}{x}dx$$
Integrating left sides:
$\int \frac{-2(v+2)}{3v^2}dv=\int \frac{1}{x}dx$$
$$-\frac{2}{3}\int \frac{1}{v}dv-\frac{4}{3}\int \frac{1}{v^2}dv=\int \frac{1}{x}dx$
$$-\frac{2}{3}\ln v +\frac{4}{3v}=\ln x +\ln C$$
where $C$ is a constant of integration.
$$\ln v^2 +\ln x^3 +\ln C=\ln(Cv^2x^3)=\frac{4}{v}$$
Subtitute when $y=vx $
$$\rightarrow \ln(Cy^2x)=\frac{4x}{y}$$
$$\rightarrow y^2x=Ce^{\frac{4x}{y}}$$
