The spherical coordinate system is orthonormal.
Proof
In spherical coordinates
\[(r, \theta , \phi)\]
we can write\[x= r sin \theta cos \phi\]
\[y= r sin \theta sin \phi\]
\[z=r cos \theta\]
Hence
\[\mathbf{r}= r sin \theta cos \phi \mathbf{i} + r sin \theta sin \phi\mathbf{j} + r cos \theta \mathbf{k}\]
Then
\[\frac{\partial \mathbf{r}}{\partial r}= sin \theta cos \phi \mathbf{i} + sin \theta sin \phi\mathbf{j} + cos \theta \mathbf{k}\]
\[\frac{\partial \mathbf{r}}{\partial \theta}= r cos \theta cos \phi \mathbf{i} + r cos \theta sin \phi\mathbf{j} - r sin \theta \mathbf{k}\]
\[\frac{\partial \mathbf{r}}{\partial \phi}= -r sin \theta sin \phi \mathbf{i} + r sin \theta cos \phi\mathbf{j} \]
\[\begin{equation} \begin{aligned} \mathbf{e_1} &= \frac{ \partial \mathbf{r} / \partial r}{| \partial \mathbf{r} / \partial r} = \frac{sin \theta cos \phi \mathbf{i} + sin \theta sin \phi\mathbf{j} + cos \theta \mathbf{k}}{ \sqrt{sin^2 \theta cos^2 \phi + sin^2 \theta sin^2 \phi + cos^2 \theta}} \\ &= sin \theta cos \phi \mathbf{i} + sin \theta sin \phi\mathbf{j} + cos \theta \mathbf{k} \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} \mathbf{e_2} &= \frac{ \partial \mathbf{r} / \partial \theta}{| \partial \mathbf{r} / \partial \theta}= \frac{r cos \theta cos \phi \mathbf{i} + r cos \theta sin \phi\mathbf{j} - r sin \theta \mathbf{k}}{ \sqrt{r^2 cos^2 \theta cos^2 \phi+ r^2 cos^2 \theta sin^2 \phi + sin^2 \theta}} \\ &= - r cos \theta cos \phi \mathbf{i} + r cos \theta sin \phi\mathbf{j} - r sin \theta \mathbf{k} \end{aligned} \end{equation}\]
\[\mathbf{e_3} = \frac{ \partial \mathbf{r} / \partial z}{| \partial \mathbf{r} / \partial z |}= \frac{-r sin \theta sin \phi \mathbf{i} + r sin \theta cos \phi \mathbf{j}}{r sin \theta} = - sin \phi \mathbf{i} + cos \phi \mathbf{j} \]
Then
\[\mathbf{e_1} \cdot \mathbf{e_1} = \mathbf{e_2} \cdot \mathbf{e_2} = \mathbf{e_3} \cdot \mathbf{e_3}=1\]
and
\[\mathbf{e_1} \cdot \mathbf{e_2} = \mathbf{e_1} \cdot \mathbf{e_3} = \mathbf{e_2} \cdot \mathbf{e_3}=\mathbf{e_2} \cdot \mathbf{e_1} = \mathbf{e_3} \cdot \mathbf{e_1} = \mathbf{e_3} \cdot \mathbf{e_2}=0\]