Assume a solution of the form [tex]\Psi(x,y)=C[/tex]. Taking the differential of both sides would give the ODE
[tex]\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0[/tex]
where [tex]\Psi_x=2y-8x^2[/tex] and [tex]\Psi_y=x[/tex]. The ODE is exact if [tex]\Psi_{xy}=\Psi_{yx}[/tex]. We have
[tex]\Psi_{xy}=(2y-8x^2)_y=2[/tex]
[tex]\Psi_{yx}=(x)_x=1[/tex]
so the ODE is not exact. However, multiplying both sides by [tex]x[/tex] gives
[tex](2xy-8x^3)\,\mathrm dx+x^2\,\mathrm dy=0[/tex]
and now
[tex](x\Psi_x)_y=(2xy-8x^3)_y=2x[/tex]
[tex](x\Psi_y)_x=(x^2)_x=2x[/tex]
so the ODE is now exact.
Now,
[tex]\Psi_x=2xy-8x^3\implies \Psi(x,y)=x^2y-2x^4+f(y)[/tex]
[tex]\Psi_y=x^2=x^2+f_y\implies f_y=0\implies f(y)=C[/tex]
So the ODE has solution
[tex]\Psi(x,y)=x^2y-2x^4+C=C[/tex]
or simply
[tex]\boxed{x^2y-2x^4=C}[/tex]
