Bernoulli differential equation help? We have the equation $$3xy' -2y=\frac{x^3}{y^2}$$ It is a type of Bernouli differential equation. So, since B. diff equation type is 
$$y'+P(x)y=Q(x)y^n$$
I modify it a little to:
$$y'- \frac{2y}{3x} = \frac{x^2}{3y^2}$$
$$y'-\frac{2y}{3x}=\frac{1}{3}x^2y^{-2}$$
Now I divide both sides by $y^{-2}$. What should I do now?

$$\text{We have $3xy^2 y'-2y^3 = x^3 \implies x (y^3)' - 2y^3 = x^3 \implies \dfrac{(y^3)'}{x^2} + y^3 \times \left(-\dfrac2{x^3}\right) = 1$}$$
$$\text{Now note that }\left(\dfrac1{x^2}\right)' = -\dfrac2{x^3}. \text{ Hence, we have }\dfrac{d}{dx}\left(\dfrac{y^3}{x^2}\right) = 1\implies \dfrac{y^3}{x^2} = x + c$$
$$\text{Hence, the solution to the differential equation is }\boxed{\color{blue}{y^3 = x^3 + cx^2}}$$
