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