We want to solve
[tex]3e^{3x} \frac{dy}{dx} = \frac{9x}{y^{2}} [/tex]
The ODE is separable into the form
[tex]3y^{2} dy = 9xe^{-3x} \\\\ y^{2} dy = 3xe^{-3x}dx [/tex]
Integrate.
[tex] \frac{1}{3} \int y^{2} dy = \int x e^{-3x} dx \\\\ \frac{y^{3}}{9} = -\frac{xe^{-3x}}{3} + \frac{1}{3} \int e^{-3x} dx \\\\ = - \frac{x}{3}e^{-3x} - \frac{1}{9} e^{-3x} + c [/tex]
[tex]y^{3} = -Ce^{-3x} (1+3x) \\\\ y = - ce^{-x} \sqrt[3]{1+3x} [/tex]
Answer: [tex]y = -ce^{-x} \sqrt[3]{1+3x}, \,\,\, c=constant[/tex]
