I want to prove that if $u$ is a cylindrically symmetric vector field in $\mathbb R^3$, then $$\frac{\partial u_x}{\partial x}=\frac{\partial u_y}{\partial y}$$
I've tried this by direct derivation, but I think I'm messing up with the unit vectors. by cylindrically symmetric I mean that $u$ is a function of $(r,z)$ only, and that it can be rotated around the z-axis
EDIT: What was here previously was wrong.
Suppose we have a cylindrically symmetric vector field $\mathbf u$, symmetric about the $z$ axis.
Then we can write, with respect to cylindrical polar basis vectors, $$\mathbf u = f(r,z)\mathbf e_r + g(r,z)\mathbf e_z.$$ Now, we have $\frac{\partial \mathbf e_z}{\partial x} = 0$ and the same for $y$. The components of $\mathbf u$ in the $x$ and $y$ directions are: $$\mathbf u_x = f(r,z)\cos\phi,\qquad\qquad\mathbf u_y = f(r,z)\sin\phi,$$ where $\phi$ is the polar angle.
As $x=r\cos\phi$, $1 = \frac{\partial r}{\partial x}\cos\phi - r\frac{\partial\phi}{\partial x}\sin\phi.$ Also, $r=\sqrt{x^2+y^2}$ so $\frac{\partial r}{\partial x} = \frac{x}{r}$
Therefore, \begin{align}\frac{\partial \mathbf u_x}{\partial x} &= \frac xrf'(r,z)\cos\phi + \frac{\partial\phi}{\partial x}f(r,z)\sin\phi\\ &=\frac{x^2}{r^2}f'(r,z) - \left(\frac1{r}-\frac x{r^2}\cos\phi\right)f(r,z)\\ &=\frac{x^2}{r^2}f'(r,z) - \frac{r-x\cos\phi}{r^2}f(r,z)\\ &=\frac1{r^2}\left(x^2f'(r,z) - \frac{y^2}{r}f(r,z)\right) \end{align} This is not the same as $\frac{\partial \mathbf u_y}{\partial y}$