Given that,

$\left(x^2+xy\right)y^1=y^2$

$\left(x^2+xy\right)\frac{dy}{dx}=y^2$
$\frac{dy}{dx}=\frac{y^2}{x^2+xy} \dots \left(1\right)$

Let $y=vx$

Differentiating w.r.t. $\text{'}x\text{'}$?

$\frac{dy}{dx}=\frac{d}{dx}\left(vx\right)$

$\frac{dy}{dx}=v+x·\frac{dv}{dx}$

$\frac{y^2}{x^2+xy}=v+x·\frac{dv}{dx}$

$\frac{v^2{x}^2}{x^2+x\left(vx\right)}=v+x\frac{dv}{dx}$

$\frac{v^2}{1+v}=v+x·\frac{dv}{dx}$

$\frac{v^2}{1+v}-v=x\frac{dv}{dx}$

$\frac{v^2-v-v^2}{1+v}=x\frac{dv}{dx}$

$\frac{-v}{1+v}=x\frac{dv}{dx}$

$\frac{1}{x}dx=\frac{-\left(1+v\right)}{v}·dv$

Integrating on both sides

$\int \frac{1}{x}dx=-\int \frac{(1+v)}{v}dv$

$\log x=-\left(\int \frac{1}{v}dv+\int \frac{v}{v}dv\right)$

$\log x=-\left[\log v+v\right]+\log c$

$\log x=-\left(\log \frac{y}{x}+\frac{y}{x}+\log c\right)$

$\log x+\log \frac{y}{x}+\log c=\frac{-y}{x}$

$\log \left(x·\frac{y}{x}·c\right)=\frac{-y}{x}$

$cy=e^{\frac{-y}{x}}$