The solution of y² dx + (3xy – 1)dy = 0 is

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{3}{\mathrm{y}}\mathrm{x}=\frac{1}{{\mathrm{y}}^2} \\ \Rightarrow \text{ I.F. }={\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{y}\right)\mathrm{dy}}={\mathrm{y}}^3 \end{gathered}$$

General solution is   $\mathrm{x}(\mathrm{I}.\mathrm{F}.)=\int \mathrm{Q}\left(\mathrm{y}\right)(\mathrm{I}.\mathrm{F}.)\mathrm{dy}+\mathrm{c}$ 

                       $x\left(y^3\right)=\int \frac{1}{y^2}{y}^{3}dy=\frac{y^2}{2}+c$