Exterior derivative for the electrostatics of a wormhole I am working in a manifold diffeomorphic to $\mathbb{R} \times \mathbb{S}^2$ with metric 
$$g = dr^2 + f(r)^2\left(d\theta + \sin^2\theta~d\phi\right)$$
where $f(r)$ is always positive. Now, I posit an electric field $E = e(r)dr$ with $e(r)$ a function to be determined. Maxwell's electrostatics equations are as follow.
$$
\begin{aligned}
dE &= 0 \\
d\star E &= 0
\end{aligned}
$$
The first equation is automatically satisfied by any $e(r)$. Now, at the moment I am obly able to take the Hodge dual of a form if I can express it in a orthonormal basis. So I move to a new basis:
$$dx^1 = dr,~~ dx^2 = f(r)d\theta,~~dx^3 = f(r)\sin(\theta)d\phi$$
and take the Hodge dual:
$$\star E = \star\left(e(r)dx^1\right) = e(r)~dx^2\wedge dx^3.$$
My problems start here. If I move back to the original frame to take the exterior derivative, I get an equation. If I take the derivative in this frame, I get a different one! Here are the two derivations.
Going back to the old frame:
$$\begin{align}
d\star E &= d\left(e(r)f(r)^2\sin\theta~d\theta\wedge d\phi\right) \\
&=\partial_r\left(e(r)f(r)^2\right)\sin\theta~dr\wedge d\theta\wedge d\phi\\
&= 0 \\\\
&\leftrightarrow \partial_r\left(e(r)f(r)^2\right) = 0
\end{align}$$
Staying in the orthonormal frame:
$$\begin{align}
d\star E &= d\left(e(r)~dx^2\wedge dx^3\right) \\
&=\partial_1e(r) ~ dx^1 \wedge dx^2 \wedge dx^3\\
&= 0 \\\\
&\leftrightarrow \partial_1e(r) = 0
\end{align}$$
Now, to my understanding, these are different equations, since $\partial_r = \partial_1$ as per my change of basis. Something must be wrong, but it's escaping me at the moment. Most probably it's about how I take the exterior derivative, but I seem to be following the same rule in both coordinates.
I am probably making a silly mistake so thanks in advance for the patience.
 A: The mistake is writing the orthonormal basis as differentials. Note that
$$d(dx^2) =d( f(r)d\theta\,)=f'(r)\,dr\wedge d\theta$$
which is not necessarily zero. Use a different name for the basis and everything becomes clear:
$$e^1 = dr,~~ e^2 = f(r)d\theta,~~e^3 = f(r)\sin(\theta)d\phi$$
so that
$$de^1 = 0\,\,\,\,\,\,\,\,\,de^2=\frac{f'}{f}e^1\wedge e^2\,\,\,\,\,\,\,\,\,de^3=\frac{f'}{f}\,e^1\wedge e^3+\frac{\cot{\theta}}{f}\,e^2\wedge e^3$$
Then in the orthonormal basis:
\begin{align}
d\star E &= d\left(e(r)\,e^2\wedge e^3\right) \\
&=\partial_1 e(r)\,e^1\wedge\,e^2\wedge e^3+e(r)\,de^2\wedge e^3-e(r)\,e^2\wedge de^3\\
&=\partial_1 e(r)\,e^1\wedge\,e^2\wedge e^3+e(r)\,\frac{f'}{f}e^1\wedge e^2 \wedge e^3-e(r)\frac{f'}{f}\,e^2\wedge \,e^1\wedge e^3\\
&=\left(\partial_r e(r)+2e(r)\frac{f'}{f}\right)\,e^1\wedge\,e^2\wedge e^3\\
&=\frac{1}{f^2}\partial_r\left(f^2 e(r)\right)\,e^1\wedge\,e^2\wedge e^3\\\\
&\leftrightarrow \partial_r\left(f^2 e(r)\right) = 0
\end{align}
which is the same equation you got with the other method.
