Hints on solving $y'=\frac{y}{3x-y^2}$ 
$$y'=\frac{y}{3x-y^2}$$

My attempt:
$$\frac{dy}{dx}=\frac{y}{3x-y^2}$$
$$dy\cdot(3x-y^2)=dx\cdot y$$
$$dy\cdot3x-dy\cdot y^2=dx\cdot y$$
Any direction? 
I need hints please $\color{red}{not}$ a full answer

Following the hint given by Chinny84, we write the equation as $$\dfrac{\mathrm{d}x}{\mathrm{d}y} = \frac{3x-y^2}{y} = \frac{3x}{y} -y$$
So $$\frac{\mathrm{d}x}{\mathrm{d}y} - \frac{3x}{y} = -y$$ which is a linear first order differential equation in $x$. 
Spoiler: Answer below.

Using an integrating factor $$I = \exp\left({\int \frac{-3}{y}}\, \mathrm{d}y \right) = \frac{1}{y^3}.$$

We get that $$\frac{x}{y^3}= \int \frac{1}{y^3}\cdot-y \, \mathrm{d}y = \frac{1}{y} + \mathrm{c}.$$

Multiplying through by $y^3$ yields 

$$x = \mathrm{c}y^3 + y^2$$

