 Solution

Write the differential form as $Mdx+Ndy$, with the coefficients as printed. A monomial integrating factor can be found from $\mu =x^{\alpha }{y}^{\beta }$. The exactness condition ${\partial }_y(\mu M)={\partial }_x(\mu N)$ givesSolving yields $\alpha =0,\beta =-3$. Thus, on $y\ne 0$,After multiplication, the differential form isHence the implicit solutions areThe branch $\boxed{y=0}$ was excluded by the integrating factor but satisfies the original equation and must be restored. On any graph branch with $y\ne 0$, differentiation of the boxed first integral recovers the original differential equation.