Let slope of the tangent line to a curve at any point P(x, y) be...

Given $\dfrac{dy}{dx} = \dfrac{x{y}^{2} + y}{x}$

$\dfrac{xdy - ydx}{{y}^{2}} = xdx$

$- d\left(\dfrac{x}{y}\right) = xdx$

$- \dfrac{x}{y} = \dfrac{{x}^{2}}{2} + c$

$∵$ Curve intersects the line $x + 2y = 4$ at $x = - 2 ⇒$ point of intersection is $\left(- 2, 3\right)$.

$∴$ Curve passes through \left(- 2, 3\right)$. </p><p>$⇒ \dfrac{2}{3} = 2 + c ⇒ c = - \dfrac{4}{3}$</p><p>$⇒ \dfrac{- x}{y} = \dfrac{{x}^{2}}{2} - \dfrac{4}{3}$</p><p>Now put \left(3, y\right)$ we get,

$⇒ \dfrac{- 3}{y} = \dfrac{19}{6}$

$⇒ y = \dfrac{- 18}{19}$