The ODE is exact, since
[tex](3y^2+\sec^2x)_y=6y[/tex]
[tex](6xy+y)_x=6y[/tex]
so there is a solution of the form [tex]\Psi(x,y)=C[/tex] satisfying
[tex]\Psi_x=3y^2+\sec^2x[/tex]
[tex]\Psi_y=6xy+y[/tex]
Integrating both sides of the first PDE wrt [tex]x[/tex] gives
[tex]\Psi=3xy^2+\tan x+f(y)[/tex]
and differentiating wrt [tex]y[/tex] gives
[tex]\Psi_y=6xy+y=6xy+f'(y)\implies f'(y)=y\implies f(y)=\dfrac{y^2}2+C[/tex]
Then
[tex]\Psi(x,y)=\boxed{3xy^2+\tan x+\dfrac{y^2}2=C}[/tex]
