Here is an attempt to derive the curl in cylindrical coordinates
\begin{align*} \omega =\text{Curl }\vec{V}(x,y,z) &= \nabla \times \textbf{u} =\begin{bmatrix} \frac{\partial}{\partial r} \\[0.3cm] \frac{1}{r} \frac{\partial}{\partial \phi}\\[0.3cm] \frac{\partial}{\partial z} \end{bmatrix} \times \begin{bmatrix} u_r \\ u_{\phi} \\ u_z \end{bmatrix} =\begin{bmatrix} \frac{\partial}{\partial r} \\[0.3cm] \frac{1}{r} \frac{\partial}{\partial \phi}\\[0.3cm] \frac{\partial}{\partial z} \end{bmatrix} \times \begin{bmatrix} 0 \\ u_{\phi} \\ 0 \end{bmatrix} \\ &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\[0.2cm] \frac{\partial}{\partial r} & \frac{1}{r}\frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\[0.2cm] u_r & u_{\phi} & u_z \end{pmatrix}\\ &= \hat{\imath}\big(\frac{1}{r}\frac{\partial u_z}{\partial u_{\phi}} - \frac{\partial u_\phi}{\partial z}\big) - \hat{\jmath}\big(\frac{\partial u_z}{\partial r} - \frac{\partial u_r}{\partial z}\big) + \hat{k}\big(\frac{\partial u_{\phi}}{\partial r} - \frac{1}{r}\frac{\partial u_r}{\partial \phi}\big) \\ &= \begin{bmatrix} \frac{1}{r}\frac{\partial u_z}{\partial u_{\phi}} - \frac{\partial u_\phi}{\partial z} \\[0.2cm] \frac{\partial u_z}{\partial r} - \frac{\partial u_r}{\partial z} \\[0.2cm] \frac{\partial u_{\phi}}{\partial r} - \frac{1}{r}\frac{\partial u_r}{\partial \phi} \end{bmatrix}\qquad \textbf{(1)}\\ &= \begin{bmatrix} \frac{1}{r}\frac{\partial u_z}{\partial u_{\phi}} - \frac{\partial u_\phi}{\partial z} \\[0.2cm] \frac{\partial u_z}{\partial r} - \frac{\partial u_r}{\partial z} \\[0.2cm] \frac{1}{r}\big( \frac{\partial r u_{\phi}}{\partial r} -\frac{\partial u_r}{\partial \phi}\big) \end{bmatrix} \qquad \textbf{(2)}\\ \end{align*}
This corresponds, I think to the wikipedia definition, with the appropriate screenshot:
Here is my question:
Going from (1) to (2) how can we mathematically assert/prove that $\frac{\partial u_{\phi}}{\partial r} = \frac{1}{r}\frac{\partial r u_{\phi}}{\partial r}$
Let us start from the components written in curvilinear coordinates
\begin{align} & (\operatorname{curl}\mathbf F)_1=\frac{1}{h_2h_3}\left (\frac{\partial (h_3F_3)}{\partial u_2}-\frac{\partial (h_2F_2)}{\partial u_3}\right ), \\[5pt] & (\operatorname{curl}\mathbf F)_2=\frac{1}{h_3h_1}\left (\frac{\partial (h_1F_1)}{\partial u_3}-\frac{\partial (h_3F_3)}{\partial u_1}\right ), \\[5pt] & (\operatorname{curl}\mathbf F)_3=\frac{1}{h_1h_2}\left (\frac{\partial (h_2F_2)}{\partial u_1}-\frac{\partial (h_1F_1)}{\partial u_2}\right ). \end{align}
In this case $(u_1, u_2, u_3) = (\rho, \phi, z)$, and $h_1 = 1$, $h_2 = \rho$, $h_3 = 1$.
Writing the curl as $\nabla\times \mathbf{F}$ is confusing to me because the cross product is more complicated in other coordinates than Cartesian, and it is not the definition but a notation of it. The same goes to the matrix form, if you want to write it in that way it would be:
$$\operatorname{curl} \mathbf{F} = \begin{vmatrix} h_1 \hat{\mathbf{e}}_1 &h_2 \hat{\mathbf{e}}_2 &h_3 \hat{\mathbf{e}}_3\\ \frac{\partial}{\partial u_1} & \frac{\partial}{\partial u_2} & \frac{\partial}{\partial u_3}\\ h_1 F_1 & h_2 F_2 & h_3 F_3 \end{vmatrix}\, .$$