$\displaystyle \dfrac{dy}{dx}=\dfrac{y}{3x-y^2}$

$\displaystyle \dfrac{dy}{dx}(3x-y^2) = y$

$\displaystyle (3x-y^2)dy = ydx$

$\displaystyle (3x-y^2)dy-ydx=0$

now this is in the form of what my professor calls "exact" except we still need an integrating factor. They are quite difficult to find, for me, and honestly i cannot get this one to work out so im sorry but, i dont think i can help.