[tex]\underbrace{(3x+6y)}_{M(x,y)}\,\mathrm dx+\underbrace{(6x-8y^3)}_{N(x,y)}\,\mathrm dy=0[/tex]
The ODE is exact if [tex]\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}[/tex]. We have
[tex]\dfrac{\partial M}{\partial y}=6[/tex]
[tex]\dfrac{\partial N}{\partial x}=6[/tex]
so the equation is indeed exact.
We're looking for a solution of the form
[tex]\Psi(x,y)=C[/tex]
By the chain rule, we have
[tex]\dfrac{\partial\Psi}{\partial x}+\dfrac{\partial\Psi}{\partial y}\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\iff\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0[/tex]
which means [tex]\dfrac{\partial\Psi}{\partial x}=M[/tex] and [tex]\dfrac{\partial\Psi}{\partial y}=N[/tex].
Now
[tex]\dfrac{\partial\Psi}{\partial x}=M\implies\displaystyle\int\frac{\partial\Psi}{\partial x}\,\mathrm dx=\int M\,\mathrm dx[/tex]
[tex]\implies\Psi(x,y)=\dfrac32x^2+6xy+f(y)[/tex]
Differentiating with respect to [tex]y[/tex] yields
[tex]\dfrac{\partial\Psi}{\partial y}=N\implies 6x+f'(y)=6x-8y^3[/tex]
[tex]\implies f'(y)=-8y^3\implies f(y)=\displaystyle\int(-8y^3)\,\mathrm dy=-2y^4+C[/tex]
and so the general solution is
[tex]\Psi(x,y)=\dfrac32x^2+6xy-2y^4+C=C[/tex]
[tex]\implies\Psi(x,y)=\dfrac32x^2+6xy-2y^4=C[/tex]
