Curl without integrals

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I have a hard time following the integral derivation of a curl in a given coordinate system and it seems somewhat meandering from the entities involved. I came up with this method, I'm wondering if someone can tell me if it can go wrong.

Suppose the line element is $\vec{ds}=\sum_i h_i dx_i \hat{e_i}$

For example in spherical coordinates. $\vec{ds}=dr\hat{r}+r\sin \phi d\theta \hat{\theta} + rd\phi \hat{\phi}$, making $h_r=1, h_\theta = r \sin \phi, h_\phi = r$

Requiring $\nabla f \cdot \vec{ds}=df$ and using the change rule, it follows that $\nabla f = \sum_i \frac{1}{h_i}\frac{\partial f}{\partial x_i} \hat{e_i}$

Let $\vec{E} = \sum_i E_i \hat{e_i}$

$\nabla \times \vec{E} = \sum_i (\nabla E_i) \times \hat{e_i}+\sum_i E_i (\nabla \times \hat{e_i})$

$\nabla (x_i) = \frac{1}{h_i} \hat{e_i}$

Since $\nabla \times \nabla f$=0

$0=\frac{-\nabla h_i}{h_i^2}\times \hat{e_i}+\frac{1}{h_i}(\nabla \times \hat{e_i})$

So $\nabla \times \hat{e_i} = \frac{\nabla h_i}{h_i}\times \hat{e_i}=\nabla(h_i) \times \nabla (x_i)$

We can now represent the curl in terms of gradients and basis vectors.

$\nabla \times \vec{E} = \sum_i \nabla(E_i) \times \hat{e_i} + \sum_i \frac{E_i \nabla h_i}{h_i}\times \hat{e_i}$

$\nabla \times \vec{E} = \sum_i \frac{\nabla (E_ih_i)}{h_i}\times \hat{e_i}=\sum_i \nabla(E_ih_i) \times \nabla(x_i)$

$(\nabla \times \vec{E})_a=\sum_i \epsilon_{abc}\frac{1}{h_b}\frac{\partial(h_iE_i)}{\partial x_b}\frac{1}{h_c}\frac{\partial x_i}{\partial x_c}=\sum_i \epsilon_{abc}\frac{1}{h_b}\frac{\partial(h_iE_i)}{\partial x_b}\frac{1}{h_c}\delta_{ic}$

$(\nabla \times \vec{E})_a=\epsilon_{abc}\frac{1}{h_bh_c}\frac{\partial (h_cE_c)}{\partial x_b}=\epsilon_{abc}\frac{h_a}{h_ah_bh_c}\frac{\partial (h_cE_c)}{\partial x_b}$

$$ \nabla \times \vec{E} = \frac{1}{h_ih_jh_k} \begin{vmatrix} h_i\hat{e_i} & h_j\hat{e_j} & h_k \hat{e_k} \\ \frac{\partial}{\partial x_i} &\frac{\partial}{\partial x_j} & \frac{\partial}{\partial x_k}\\ h_iE_i & h_jE_j & h_kE_k \end{vmatrix} $$

I find this easier to follow than the integral formulation. For integrals to make sense, you need pictures to get help get the surfaces right. It's easy to get the signs wrong and the role of the scaling factors are weird.

Anyone see a mistake? The final result matches what I've seen in tables elsewhere.