Calculating $ d \Phi^{*} \omega$ and $ \Phi^{*} d\omega$ 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?
 A: 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.
