<s>[INST] Solve $y'-\frac{y}{x}=\frac{(x+y)^2}{x^2}$ $$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\\\
x^2y'-xy &=x^2+2xy+y^2\\
\end{aligned}$$
I have no idea what is the next step.
Updated:
Let $y=vx$, which $y′=v+v′x$
$$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\
xv'&=(v+1)^2\\
\frac{v'}{(v+1)^2}&=\frac{1}{x}\\
\int\frac{1}{(v+1)^2}\,dv &=\int\frac{1}{x}\, dx\\
-\frac{1}{v+1}&=\ln x+C\\
v&=-\frac{\ln x+1+C}{\ln x+C}\\
y&=-\frac{x(\ln x+1+C)}{\ln x+C}
\end{aligned}$$
 [/INST] $$ax^2+bx+c=0\implies x=(-b\pm \sqrt{b^2-4ac})/(2a)$$
substitute: $ y^{(1)}=\frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)}$
solution: $y=\int \frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)} dx$
</s>