Let $\omega \in \Omega^2(\mathbb{R}^3)$ as follow: $\omega = x dy\land dz + y dz \land dx + z dx \land dy $. Let $\Phi: \Bbb R^3\to \Bbb R^3$ be given by
$$\Phi(r, \phi, \psi) = (r\cos\phi \cos\psi, r\sin \phi \cos\psi, r\sin \psi) .$$
I can not figure out how to calculate $d\Phi^{*}\omega $ and $ \Phi^{*} d\omega$.
Are they supposed to be equal?
The computation is very long and mechanical, I am only doing it for the first term, the rest is left for you to try.
$$\Phi^* (x \,dy\wedge dz) = (x\circ\Phi)\big|_{(r,\phi,\psi)} d(y\circ\Phi)\big|_{(r,\phi,\psi)}\wedge d(z\circ\Phi)\big|_{(r,\phi,\psi)}.\tag{1}$$ First term: $$ (x\circ\Phi)\big|_{(r,\phi,\psi)} = r\cos\psi \cos\phi. $$ Second term: $$ d(y\circ\Phi)\big|_{(r,\phi,\psi)} = d(r\cos\psi \sin\phi) \\ = \cos\psi \sin\phi\, dr - r\sin\psi \sin\phi\, d\psi + r\cos\psi \cos\phi\,d\phi. $$ Third term: $$ d(z\circ\Phi)\big|_{(r,\phi,\psi)} = d(r\sin\psi) = \sin\psi\,dr + r\cos\psi\,d\psi. $$ Therefore (1) is: $$ (r\cos\psi \cos\phi)(\cos\psi \sin\phi\, dr - r\sin\psi \sin\phi\, d\psi + r\cos\psi \cos\phi\,d\phi)\wedge (\sin\psi\,dr + r\cos\psi\,d\psi), $$ make use of $dr\wedge dr = 0$ and $d\psi\wedge d\psi = 0$, $dr\wedge d\psi = -d\psi\wedge dr$ we have: $$ \Phi^* (x \,dy\wedge dz) = \\ r^2 \cos\psi \cos\phi \sin\phi \,dr\wedge d\psi \\ + r^2 \cos^2\psi \sin \psi \cos^2 \phi \,dr\wedge d\phi \\ + r^3 \cos^3\psi \cos^2\phi \,d\phi\wedge d\psi. $$ When taking exterior derivative, notice $$ d( \cos\psi \cos\phi \sin\phi \,dr\wedge d\psi) = \\ \color{\red}{ \frac{\partial }{\partial r}\cos\psi \cos\phi \sin\phi \,dr\wedge (dr\wedge d\psi)} \\ + \color{\red}{\frac{\partial }{\partial \psi}\cos\psi \cos\phi \sin\phi \,d\psi\wedge (dr\wedge d\psi) } \\ + \color{\green}{\frac{\partial }{\partial \phi}\cos\psi \cos\phi \sin\phi \,d\phi\wedge (dr\wedge d\psi)}, $$ red terms are gone, only green term left. $$ d\Phi^* (x \,dy\wedge dz) = \\ \frac{\partial }{\partial \phi} r^2 \cos\psi \cos\phi \sin\phi \,d\phi\wedge (dr\wedge d\psi) \\ + \frac{\partial }{\partial \psi}r^2 \cos^2\psi \sin \psi \cos^2 \phi \,d\psi\wedge(dr\wedge d\phi) \\ + \frac{\partial }{\partial r}r^3 \cos^3\psi \cos^2\phi \,dr\wedge(d\phi\wedge d\psi). $$ Then we can combine these terms by: $$ dr \wedge d\psi\wedge d\phi = d\phi\wedge (dr\wedge d\psi) = -d\psi\wedge(dr\wedge d\phi) =-dr\wedge(d\phi\wedge d\psi). $$ Do the same for the other two terms (left for you to try): $$ d\Phi^* \omega = d\Phi^* (x \,dy\wedge dz) + d\Phi^* (y \,dz\wedge dx) + d\Phi^* (z \,dx\wedge dy). $$
For $d\omega = 3\,dx\wedge dy\wedge dz$, we can use (1) again: $$\Phi^* d\omega = d(x\circ\Phi)\big|_{(r,\phi,\psi)} \wedge d(y\circ\Phi)\big|_{(r,\phi,\psi)}\wedge d(z\circ\Phi)\big|_{(r,\phi,\psi)}.$$ The computation is similar as above.