[tex]\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0[/tex]
For the ODE to be exact, we require that [tex]M_y=N_x[/tex], which we'll verify is not the case here.
[tex]M_y=21xy^2+6[/tex]
[tex]N_x=2xy^2-1[/tex]
So we distribute an integrating factor [tex]i(x,y)[/tex] across both sides of the ODE to get
[tex]iM\,\mathrm dx+iN\,\mathrm dy=0[/tex]
Now for the ODE to be exact, we require [tex](iM)_y=(iN)_x[/tex], which in turn means
[tex]i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM[/tex]
Suppose [tex]i(x,y)=x^ry^s[/tex]. Then substituting everything into the PDE above, we have
[tex]x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)[/tex]
[tex]19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s[/tex]
[tex]19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s[/tex]
[tex]\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2[/tex]
so that our integrating factor is [tex]i(x,y)=x^5y^{-2}[/tex]. Our ODE is now
[tex](7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0[/tex]
Renaming [tex]M(x,y)[/tex] and [tex]N(x,y)[/tex] to our current coefficients, we end up with partial derivatives
[tex]M_y=7x^6-6x^5y^{-2}[/tex]
[tex]N_x=7x^6-6x^5y^{-2}[/tex]
as desired, so our new ODE is indeed exact.
Next, we're looking for a solution of the form [tex]\Psi(x,y)=C[/tex]. By the chain rule, we have
[tex]\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)[/tex]
Differentiating with respect to [tex]y[/tex] yields
[tex]\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}[/tex]
[tex]\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C[/tex]
Thus the solution to the ODE is
[tex]\Psi(x,y)=x^7y+x^6y^{-1}=C[/tex]
