Let slope of the tangent line to a curve at any point $$P(x, y)$$ be given by $$\frac{xy^2 + y}{x}$$. If the curve intersects the line $$x + 2y = 4$$ at $$x = -2$$, then the value of $$y$$, for which the point $$(3, y)$$ lies on the curve, is:

The slope of the tangent at $$(x, y)$$ is $$\dfrac{dy}{dx} = \dfrac{xy^2 + y}{x} = y^2 + \dfrac{y}{x}$$.

This is a Bernoulli equation. Dividing through by $$y^2$$: $$\dfrac{1}{y^2}\dfrac{dy}{dx} - \dfrac{1}{xy} = 1$$.

Let $$v = \dfrac{1}{y}$$, so $$\dfrac{dv}{dx} = -\dfrac{1}{y^2}\dfrac{dy}{dx}$$. The equation becomes $$-\dfrac{dv}{dx} - \dfrac{v}{x} = 1$$, i.e., $$\dfrac{dv}{dx} + \dfrac{v}{x} = -1$$.

The integrating factor is $$e^{\int dx/x} = x$$. Multiplying: $$\dfrac{d}{dx}(vx) = -x$$.

Integrating: $$vx = -\dfrac{x^2}{2} + C$$, i.e., $$\dfrac{x}{y} = -\dfrac{x^2}{2} + C$$.

To find $$C$$, the curve intersects $$x + 2y = 4$$ at $$x = -2$$. From the line: $$-2 + 2y = 4$$, so $$y = 3$$. Substituting $$(-2, 3)$$: $$\dfrac{-2}{3} = -\dfrac{4}{2} + C = -2 + C$$, giving $$C = -\dfrac{2}{3} + 2 = \dfrac{4}{3}$$.

The curve equation is $$\dfrac{x}{y} = -\dfrac{x^2}{2} + \dfrac{4}{3}$$. At $$x = 3$$: $$\dfrac{3}{y} = -\dfrac{9}{2} + \dfrac{4}{3} = \dfrac{-27 + 8}{6} = -\dfrac{19}{6}$$.

Therefore $$y = \dfrac{3 \times 6}{-19} = -\dfrac{18}{19}$$.