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