One short question:
If $\Phi\colon\mathbb{R}^3\to\mathbb{R}^3$, defined by $$ \begin{pmatrix}r\\\vartheta\\\phi\end{pmatrix}\mapsto\begin{pmatrix}r\sin \vartheta\cos \phi\\r\sin \vartheta \sin\phi\\r\cos\vartheta\end{pmatrix}, $$ what are then the exterior products $$ d\Phi_1\wedge d\Phi_2,~~~~~~~~~~d\Phi_1\wedge d\Phi_3,~~~~~~~~~~d\Phi_2\wedge d\Phi_3? $$
Hint:
$$d\Phi_1:=\frac{\partial \Phi_1}{\partial r}dr+\frac{\partial \Phi_1}{\partial \theta}d\theta+\frac{\partial \Phi_1}{\partial \phi}d\phi= \sin\theta\cos\phi dr+r\cos\theta\cos\phi d\theta-r\sin\theta \sin\phi d\phi,$$
and similarly for the other components $\Phi_2$, $\Phi_3$ of $\Phi$, i.e.
$$d\Phi_2:=\frac{\partial \Phi_2}{\partial r}dr+\frac{\partial \Phi_2}{\partial \theta}d\theta+\frac{\partial \Phi_2}{\partial \phi}d\phi= \sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi,$$
$$d\Phi_3:=\frac{\partial \Phi_3}{\partial r}dr+\frac{\partial \Phi_3}{\partial \theta}d\theta+\frac{\partial \Phi_3}{\partial \phi}d\phi= \cos\theta dr-r\sin\theta d\theta,$$
as $\frac{\partial \Phi_3}{\partial \phi}=\frac{\partial r\cos\theta}{\partial \phi}=0.$
To arrive at the answer you need to remember that, by definition of wedge product
$$dx_i\wedge dx_j=-dx_j\wedge dx_i,$$
for all $i,j=1,2,3$ with $x_1=r$, $x_2=\theta$ and $x_3=\phi$. In particular
$$dr\wedge dr=d\theta\wedge d\theta=d\phi\wedge d\phi=0.~(*)$$
Moreover $f(r,\theta,\phi)dx_i\wedge g(r,\theta,\phi)dx_j:= f(r,\theta,\phi)g(r,\theta,\phi)dx_i\wedge dx_j$, for all functions $f$ and $g$.
Let us compute $d\Phi_1\wedge d\Phi_2$: using the above formulae we have
$$d\Phi_1\wedge d\Phi_2=(\sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi)\wedge(\sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi); $$
the wedge product $\wedge$ is linear: using the property $(*)$ we collect only the non trivial contributions arriving at
$$d\Phi_1\wedge d\Phi_2=r\sin\theta\sin\phi \cos\theta\sin\phi dr\wedge d\theta+ r \sin^2\theta\sin\phi \cos\phi dr\wedge d\phi+\\ r\cos\theta \sin\theta \sin^2\phi d\theta\wedge dr+ r^2\cos\theta r\sin\theta \sin\phi \cos\phi d\theta\wedge d\phi+\\ r\sin^2\theta\cos\phi\sin\phi d\phi\wedge dr + r^2\sin\theta\cos\theta\cos\phi\sin\phi d\phi \wedge d\theta; $$
using $dr\wedge d\theta=-d\theta\wedge dr$ etc...we arrive at the final result, i.e.
$$d\Phi_1\wedge d\Phi_2=(r\sin\theta\sin\phi \cos\theta\sin\phi-r\cos\theta \sin\theta \sin^2\phi) dr\wedge d\theta+ (r \sin^2\theta\sin\phi \cos\phi-r\sin^2\theta\cos\phi\sin\phi ) dr\wedge d\phi+\\ (r^2\cos\theta r\sin\theta \sin\phi \cos\phi-r^2\sin\theta\cos\theta\cos\phi\sin\phi ) d\theta\wedge d\phi. $$