nabla in spherical coordinates using tensor transformation rule

Question

Usually people derive nabla coordinates in spherical system using orthogonal basis or another tricks, I just want to apply general rule for $\left(\begin{array}[c] 00 \\ 1\end{array}\right)$ tensor. As I understand, if \begin{equation} d f = \dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy+\dfrac{\partial f}{\partial z}dz \end{equation} Then in cartesian coordinates \begin{equation} \nabla_i = \dfrac{\partial f}{\partial x_i} \end{equation} for $\left(\begin{array} 00 \\ 1\end{array}\right)$ tensor coordinates after changing atlas from $X\rightarrow Y$: \begin{equation} \begin{array}[c] xx = r\sin(\theta)\cos(\varphi)\\ y =r\sin(\theta)\sin(\varphi)\\ z=r\cos(\theta) \end{array}\Rightarrow \begin{array}[c] rr = \sqrt{x^2+y^2+z^2}\\ \theta = arctan(\sqrt{x^2+y^2}/z)\\ \varphi = arctan(y/x) \end{array} \end{equation} have to transform like: \begin{equation} \nabla_k =\nabla_i\dfrac{\partial x^i}{\partial y^k} \end{equation} So \begin{equation} \nabla_r = \dfrac{\partial f}{\partial x_i}\dfrac{\partial x^i}{\partial r}=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial r}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial r}= \end{equation} \begin{equation} =\dfrac{\partial f}{\partial x}\sin(\theta)\cos(\varphi)+\dfrac{\partial f}{\partial y}\sin(\theta)\sin(\varphi)+\dfrac{\partial f}{\partial z}\cos(\theta) \end{equation} but wikipedia says: \begin{equation} \nabla_r = \dfrac{\partial }{\partial r} \end{equation}