Partial derivative coordinates of tangent vectors and covectors

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I am learning differential geometry and it has triggered a more general question about the partial derivatives, as they relate to the coordinate functions of vectors and covectors.

Using the following for vectors $${\partial \over \partial f } = \sum a^i {\partial \over \partial x^i }$$

where $$a^i = {\partial x^i \over \partial f}$$

and for covectors $$df = \sum b^i dx^i $$

where $$ b^i = {\partial f \over \partial x^i}$$

Now using the cylindrical coordinate frame as an example with $x = r \cos \theta$ and $y = r \sin \theta$ we find $$dr = \cos \theta dx + \sin \theta dy $$

so $ {\partial r \over \partial x} = \cos \theta $

and since $x = r \cos \theta $ we also have $ {\partial x \over \partial r} = \cos \theta $

Doing the same for $y$ we get the same coordinate functions in the vector and in the covector

$${\partial \over \partial r } = \cos \theta {\partial \over \partial x } + \sin \theta {\partial \over \partial y } $$

looking at $d\theta$ however, we get that

$${\partial \theta \over \partial x} = - {1\over r} \sin \theta $$ $${\partial \theta \over \partial y} = {1\over r} \cos \theta $$

while

$${\partial x \over \partial \theta} = - r \sin \theta$$ $${\partial y \over \partial \theta} = r \cos \theta$$

... leading to the unit vector ${1 \over r}{\partial \over \partial \theta}$ and its covector $r d\theta$, which similarly share the same coordinate functions

Now, the question... what can we say about the relationship between $\partial f \over \partial x$ and $\partial x \over \partial f$ ? Specifically I am looking for a condition in which the coordinates of a vector and covector are equal as in the attitude matrix of a frame field of orthonormal vectors. I am able to show that if vectors and covectors have the same coordinates that they will be unit vectors, but I am looking for the reverse proof, the condition on which vectors and covectors will have the same coordinates $a^i = b^i$

My own attempt so far is as follows, considering coordinate functions constant and integrating $df$ ...

$$f = \sum b^i x^i + c $$

$$df ({\partial \over \partial f} ) = {\partial \over \partial f} (f) = (\sum a^i {\partial \over \partial x^i}) \circ (\sum b^j x^j + c) $$

and presuming ${\partial x^i \over \partial x^j} = \delta _{ij}$ then

$$ df ({\partial \over \partial f} ) = \sum a^i b^i $$ and of course

$$ df ({\partial \over \partial f} ) = 1$$

so, $\sum a^i b^i = 1$

Now $ ||E_1||^2 = \sum a^i a^i $ and if the coordinates are equal, then $a^i = b^i$ so $\sum a^i a^i = 1$ so we have a unit vector / covector.

So unit length is a condition for the vector and covector coordinates to be the same, but is this condition also sufficient?