## Calculus (3rd Edition)

$$\frac{dy}{dx}=\frac{y+2x}{1-x}, \quad x\neq 1.$$
Simplify the given equation first by multiplying both sides by $x$; We then have $y=x^2+xy$. Now, by differentiating the equation $y=x^2+xy$ with respect to $x$, we get $$\frac{dy}{dx}=2x+y+x \frac{dy}{dx}\Longrightarrow \frac{dy}{dx}(1-x)=y+2x$$ and hence $$\frac{dy}{dx}=\frac{y+2x}{1-x}, \quad x\neq 1.$$