Considering the general expression of gradient with the directional derivative operator on $f$ function along $\vec{v}$ vector :
$$df(v)=\langle\text{grad}(f),v\rangle = g^{ij}\partial_{i} f v_{j}=\partial_{i} f v^{i}$$
taking $\partial_{i} = \dfrac{\partial}{\partial x^{i}}$ (with $x^{i}$ contravariant coordinates).
Can I write also :
$$df(v)=\langle\text{grad}(f),v\rangle = \partial^{i} f v_{i}$$
with $\partial^{i} = g^{ij} \partial_{j} =\dfrac{\partial}{\partial x_{i}}$ where $x_{i}$ are covariant coordinates ??
i.e, I don't know if I can raise up the index of $\partial_{j}$ multiplying it by $g^{ij}$ while defining $\partial^{i} = \dfrac{\partial}{\partial x_{i}}$ ?
Regards
UPADTE 1 : it may be that I do confusions between covariant/contravariant coordinates of a vector and curvilinear coordinates (curviliear coordinates don't have covariant/contravariants properties)
UPDATE 2 : I am stucked on the following expression :
$$\vec{e_{i}}=g_{ij}\vec{e^{j}} = \partial_{i}=g_{ij} \partial^{j} = = g_{ij}dx^{j} = \dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} \neq \dfrac{\partial}{\partial x^{i}}\quad(1)$$
In the same time, I can write :
$$\vec{e_{i}}=g_{ij}\vec{e^{j}}=\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j}=\delta_{ij} \dfrac{\partial}{\partial x^{j}} = \dfrac{\partial}{\partial x^{i}}\quad(2)$$
What's one of these two above expressions, (1) and (2), is correct ?
The main issue is that I don't know if I can express the basis vector $\vec{e_{i}}$ as a linear combination of 1-forms $\vec{e^{i}}=\partial/\partial x^{i}$ (dual basis vectors) : at first sight, it doesn't make any sense. Anyone could help me with this contradiction ?