Doubt about the use of partial derivatives: what's the solution? I should calculate this mixed product, where $\overline r=\overrightarrow{OP}$,
$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi}),$$
to get with the determinant,
$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi})=\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta}  \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}=\color{red}{r^3\sin \theta}.\tag 1$$
If 
$$\begin{cases}  
x=r(\theta,\varphi)\sin \theta\cos \varphi\\ 
y=r(\theta,\varphi)\sin \theta\sin \varphi\\
z=r(\theta,\varphi)\cos \theta
\end{cases}$$
why are they true these systems?
$$\begin{cases}  
x_\theta(\theta,\varphi)=r_\theta\sin \theta\cos\varphi+r\cos \theta\cos \varphi\\ 
x_\varphi(\theta,\varphi)=r_\varphi\sin \theta\cos\varphi-r\sin\theta\sin \varphi 
\end{cases} \tag 2$$
$$\begin{cases}  
y_\theta(\theta,\varphi)=r_\theta\sin \theta\sin\varphi+r\cos \theta\sin \varphi\\ 
y_\varphi(\theta,\varphi)=r_\varphi\sin \theta\sin\varphi+r\sin\theta\cos \varphi
\end{cases} \tag 3$$
$$\begin{cases}  
z_\theta(\theta,\varphi)=r_\theta\cos\theta-r\sin \theta\\ 
z_\varphi(\theta,\varphi)=r_\varphi\cos\theta
\end{cases} \tag 4$$
With $(2), (3), (4)$ I have not get $r^3\sin \theta$ with the determinant $(1)$. 
Please, can I have your attention and help?
 A: A strategy suggestion to calculate by hand this determinant:
First, I'll use condensed notations: $r$ instead of $r(\theta,\varphi)$. Next, observe each column is the sum of two column vectors
$$\begin{bmatrix}x\\x_\theta\\x_\varphi\end{bmatrix}=\underbrace{\begin{bmatrix}r\sin\theta\cos\varphi\\r_\theta\sin\theta\cos\varphi\\r_\varphi\sin\theta\cos\varphi\end{bmatrix}}_{\textstyle C_x}+\underbrace{\begin{bmatrix}0\\r\cos\theta\cos\varphi\\-r\sin\theta\sin\varphi\end{bmatrix}}_{\textstyle D_x}=\sin\theta\cos\varphi\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}+r\begin{bmatrix}0\\\cos\theta\cos\varphi\\-\sin\theta\sin\varphi\end{bmatrix}$$
and similarly for the other columns.
Now the determinant, with these notations becomes, by the multilinearity properties of  determinants
\begin{align}\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta}  \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}&=\begin{vmatrix}C_x{+}D_x & C_y {+}D_y& C_z{+}D_z\\ \end{vmatrix}\\&=\bigl|C_x\enspace C_y\enspace C_z \bigr|+\bigl|D_x\enspace  C_y\enspace C_z \bigr|+\bigl|C_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace C_y\enspace D_z \bigr|\\[1ex]
&\quad+\bigl| C_x\enspace D_y\enspace C_z \bigr|+\bigl|D_x\enspace D_y\enspace C_z \bigr|+\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace D_z \bigr|
\end{align}
Each of $C_x, C_y, C_z$ is collinear to the column vector $\;\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}$, so any of the determinants in the sum which comprises two of these column vectors is $0$, further $\bigl| D_x\enspace D_y\enspace D_z \bigr|$ has a row of $0$s. Therefore there remains
$$\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta}  \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}
=\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl|D_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace C_z \bigr|,
$$
which can easily be computed by hand.
Hope this will help!
A: Equations (2), (3) and (4) are true because of the product rule for derivatives. 
A: Hint: write out the determinant in the matrix form.
Step 1: Multiply the first row by $-r_{\theta}/r$ and add it to the second row. Do similar thing with respect to the third row. You will see that all the terms involving $r_{\theta}$ and $r_{\phi}$ will be canceled.
Step 2: Extract the factors $r,r,r\sin\theta$ outside from the first, second and third row, respectively. This explains the factor $r^3\sin\theta$.
Step 3: Work on the remaining determinant by expanding with respect to the third row. It can be easily checked that the result is $\sin^2\phi+\cos^2\phi=1$, hence the result
