I'm trying to integrate the 2-form $\omega = A(y) dx \wedge dy - dx \wedge dz + B(z)dz \wedge dy $ in the set $R_f=\{(x,y,z),\quad z=f(x,y)\quad x^2+y^2 \neq 1 \}$ with $f$ a differentiable function from $\mathbb{R}^2$ to $\mathbb{R}$ such that $f(x,y)=0$ for $(x,y)$ such that $x^2+y^2=1$.
To do this I found $\theta = A(y)x dy + z dx - yB(z)dz$ such that $d\theta = \omega$ and applied Stokes' theorem getting that:
$$\int_{R_f} \omega = \int_{\partial R_f} \theta$$
I tried then to expand the expression that I got and compute the integral but I couldn't solve it. I've been trying to prove that with these conditions $dz=0$ and then I would be able to solve the problem.
How should I proceed?
You're doing great so far. Let me use the name $\eta$ for the thing you call $\theta$, and then continue.
You need to integrate $\eta$ over something that projects to the unit circle in the $xy$ plane. So let's take $$ \gamma(t) = (\cos t, \sin t, f(\cos t, \sin t)) $$ which parameterizes the set $\partial R_f$ nicely, and use it to compute a path integral. You need to evaluate \begin{align} \int_0^{2\pi} \eta(\gamma(t))(\gamma'(t)) dt &= \int \left. (A(y)x dy + z dx - yB(z)dz)\right|_{x = \cos t, y = \sin t, z = f(\cos t, \sin t)} [-\sin t, \cos t, f_x(\cos t, \sin t)(-sin t) + f_y (\cos t, \sin t) (\cos t)] dt\\ &= \int (A(\sin t) \cos t dy + f(\cos t, \sin t) dx - \sin t B(f(\cos t, \sin t)) dz))[-\sin t, \cos t, f_x(\cos t, \sin t)(-sin t) + f_y (\cos t, \sin t) (\cos t)] dt\\ &= \int A(\sin t) \cos t \cos t + f(\cos t, \sin t) (-\sin t) - \sin t B(f(\cos t, \sin t)) (f_x(\cos t, \sin t)(-\sin t) + f_y (\cos t, \sin t) (\cos t))~ dt\\ &= \int A(\sin t) \cos^2 t -\sin t ~f(\cos t, \sin t) - \sin t ~B(f(\cos t, \sin t)) (f_x(\cos t, \sin t)(-\sin t) + f_y (\cos t, \sin t) (\cos t))~ dt \end{align}
That last integral's not pretty, but it's what you need to look at.