It's my first time learning differential forms and for that I am using Advanced Calculus from C.H. Edwards. Indeed, here are the definitions so far:
$\omega = Pdx + Qdy$ is a 1-form, $d\omega = (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dxdy$ is a 2-form. And, if $F:\Bbb R^2\to \Bbb R^2$ is a one to one function whose Jacobian is positive at every point, then define the pullback of $\omega$ and $\alpha$, where $\alpha = g(x,y)dxdy$ is a 2-form, as follows $$F^* \omega = P(F)(\frac{\partial F_1}{\partial u}du + \frac{\partial F_1}{\partial v}dv) + Q(F)(\frac{\partial F_2}{\partial u}du + \frac{\partial F_2}{\partial v}dv)$$ and $$F^*\alpha = (g\circ F)(det\ Jacobian(F))dudv.$$
So, using the above definitions, I need to prove that $F^* d\omega = dF^*\omega$. Indeed, after doing some computations, I got that $$dF^*\omega = (\frac{\partial P\circ F}{\partial u}\frac{\partial F_1}{\partial v} + \frac{\partial Q\circ F}{\partial u}\frac{\partial F_2}{\partial v} - \frac{\partial P\circ F}{\partial v}\frac{\partial F_1}{\partial u} - \frac{\partial Q\circ F}{\partial v}\frac{\partial F_2}{\partial u})dudv = (\frac{\partial P}{\partial x}\frac{\partial F_1}{\partial u}\frac{\partial F_1}{\partial v} + \frac{\partial P}{\partial y}\frac{\partial F_2}{\partial u}\frac{\partial F_1}{\partial v}-\frac{\partial P}{\partial x}\frac{\partial F_1}{\partial v}\frac{\partial F_1}{\partial u}-\frac{\partial P}{\partial y}\frac{\partial F_2}{\partial v}\frac{\partial F_1}{\partial u} + \frac{\partial Q}{\partial x}\frac{\partial F_1}{\partial u}\frac{\partial F_2}{\partial v} + \frac{\partial Q}{\partial y}\frac{\partial F_2}{\partial u}\frac{\partial F_2}{\partial v}-\frac{\partial Q}{\partial x}\frac{\partial F_1}{\partial v}\frac{\partial F_2}{\partial u}-\frac{\partial Q}{\partial y}\frac{\partial F_2}{\partial v}\frac{\partial F_2}{\partial u})dudv = (\frac{\partial P}{\partial y}\frac{\partial F_2}{\partial u}\frac{\partial F_1}{\partial v}-\frac{\partial P}{\partial y}\frac{\partial F_2}{\partial v}\frac{\partial F_1}{\partial u} + \frac{\partial Q}{\partial x}\frac{\partial F_1}{\partial u}\frac{\partial F_2}{\partial v}-\frac{\partial Q}{\partial x}\frac{\partial F_1}{\partial v}\frac{\partial F_2}{\partial u})dudv= (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})(\frac{\partial F_1}{\partial u}\frac{\partial F_2}{\partial v} - \frac{\partial F_1}{\partial v}\frac{\partial F_2}{\partial u})dudv,$$ but I'm stuck, since $$F^*d\omega = ((\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\circ F)(\frac{\partial F_1}{\partial u}\frac{\partial F_2}{\partial v} - \frac{\partial F_1}{\partial v}\frac{\partial F_2}{\partial u})dudv$$ and from there I don't know how to get the expression I got in $dF^*\omega$.