Answer
$y=x\tan (\ln|x|+C)$
Work Step by Step
Given:
$$\frac{dy}{dx}=\frac{y^2+xy+x^2}{x^2}$$
$$\frac{dy}{dx}=(\frac{y}{x})^2+\frac{y}{x}+1$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$v+x\frac{dv}{dx}=v^2+v+1$$
$$x\frac{dv}{dx}=v^2+1$$
$$x\frac{dv}{v^2+1}=dx$$
$$\frac{dv}{v^2+1}=\frac{dx}{x}$$
Integrating left side:
$$\int \frac{dv}{v^2+1}=\int \frac{dx}{x}+C$$
$$\arctan v=\ln|X| + C$$
$$\tan v=\tan (\ln|x|+C)$$
Subtitute when $y=vx \rightarrow y=x\tan (\ln|x|+C)$
