Trying to get practice with differential forms and related topics I had tried to calculate curl in spherical coordinates without mystic "del" operator.
Let $rot\: F = (\star d (F^\sharp))^\flat$;
$x = r \cos \phi \sin \theta$
$y = r \sin \phi \sin \theta$
$z = r \cos \theta $
Non normalized basis of tangent space $T$:
$\bar e_r = (\frac{dx}{dr}, \frac{dy}{dr}, \frac{dz}{dr}) = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$
$\bar e_\phi = (\frac{dx}{d\phi},\frac{dx}{d\phi},\frac{dz}{d\phi}) = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$
$\bar e_\theta = (\frac{dx}{d\theta},\frac{dx}{d\theta},\frac{dz}{d\theta}) = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$
Norms:
$\|\bar e_r\| = 1$
$\|\bar e_\phi\| = r \sin \theta$
$\|\bar e_\theta\| = r$
Orthonormal basis of tangent space $T$:
$ e_r = \frac{\bar e_r}{\| \bar e_r\|} = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$
$ e_\phi = \frac{\bar e_r}{\| \bar e_r\|} = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$
$ e_\theta = \frac{\bar e_r}{\| \bar e_r\|} = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$
Let $w_r, w_\phi , w_\theta$ be corresponding basis in $T^*$. Then
$dr = w_r$
$d\phi = \frac{1}{r\sin \theta} w_\phi$
$d\theta = \frac{1}{r} w_\theta$
So, going back to the $rot$:
$F = F^1 e_r + F^2 e_\phi + F^3 e_\theta$
$F^\sharp = F^1 w_r + F^2 w_\phi + F^3 w_\theta$
$d (F^\sharp) = $ $$ = F^1_\theta d\theta \wedge w_r + F^1_\phi d\phi \wedge w_r $$ $$ + F^2_\theta d\theta \wedge w_\phi + F^2_r dr \wedge w_\phi$$ $$ + F^3_\phi d\phi \wedge w_\theta + F^3_r dr \wedge w_\theta$$
$$ = F^1_\theta \frac{1}{r} w_\theta \wedge w_r + F^1_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_r $$ $$ + F^2_\theta \frac{1}{r} w_\theta \wedge w_\phi + F^2_r w_r \wedge w_\phi$$ $$ + F^3_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_\theta + F^3_r w_r \wedge w_\theta$$
$$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_r \wedge w_\phi$$ $$ + (F^3_r - F^1_\theta \frac{1}{r}) w_r \wedge w_\theta$$ $$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_\phi \wedge w_\theta$$
$\star d (F^\sharp) = $ $$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_\theta$$ $$ - (F^3_r - F^1_\theta \frac{1}{r}) w_\phi$$ $$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_r$$
$rot F = (\star d (F^\sharp))^\flat = $ $$= (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) e_r$$ $$- (F^3_r - F^1_\theta \frac{1}{r}) e_\phi$$ $$+ (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) e_\theta$$
But in wikipedia the answer is completely different. Please, help me to find which operations i had understood wrongly.
First of all, your notation for $\flat$ and $\sharp$ is reversed: if $F$ is a vector field, $F^{\flat}$ is a covector field (1-form); and if $\omega$ is a 1-form, then $\omega^{\sharp}$ is a vector field, so the correct way of writing the curl definition is $\text{curl}(F)= (\star d(F^{\flat}))^{\sharp}$.
Your mistake is in calculating the exterior derivatives. Because you expressed your 1-forms in terms of $w_r, w_{\theta},w_{\phi}$, you need to apply the product rule when calculating $d$. You can't just take the $d$ of the coefficient and wedge it. For example, if $f$ is a smooth function and you have a 1-form $\alpha=fw_{\theta} = f\cdot rd\theta$, then the way you wrote the calculation is $d\alpha = df\wedge w_{\theta} = \frac{\partial f}{\partial r}\,dr \wedge w_{\theta} + \frac{\partial f}{\partial \phi}\,d\phi \wedge w_{\theta}$; but this is wrong. The correct way to do it is \begin{align} d\alpha &= df\wedge w_{\theta} +f (dw_{\theta}) = df\wedge w_{\theta} + f\,dr\wedge d\theta \end{align} Or, another way of saying it is that $\alpha = (rf)\,d\theta$, so \begin{align} d\alpha &= \frac{\partial (rf)}{\partial r}\,dr\wedge d\theta + \frac{\partial (rf)}{\partial \phi}\,d\phi\wedge d\theta \end{align} You made this mistake for each term.
Calculating the curl is annoying for the following reason: exterior derivatives are easy to calculate when you express everything in terms of the coordinate 1-forms (i.e in terms of $dr,d\theta,d\phi$). On the other hand, when you do the musical isomorphisms, various factors will pop up, and then again when you want to use the hodge star, that is simplest when using an orthonormal coframe (i.e the $w_r,w_{\theta},w_{\phi}$ you mention). Because of this, there's a lot of factors to keep track of.
The following works for any orthogonal system of coordinates $(x^1,x^2,x^3)$ (i.e such that the metric tensor has components $g_{ij}:=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=0$ if $i\neq j$). Start by writing $F=F^1e_1 + F^2e_2 + F^3e_3$, then you'll find \begin{align} F^{\flat}&= F^1 \omega^1 + F^2\omega^2 + F^3\omega^3\\ &= (F^1\sqrt{g_{11}})\,dx^1+ (F^2\sqrt{g_{22}})\,dx^2 +(F^3\sqrt{g_{33}})\,dx^3 \end{align} Or, if you define $f_i:= F^i\cdot \sqrt{g_{ii}}=F^ih_i$ (no summation convention) then $F^{\flat}=\sum_{i=1}^3f_i\,dx^i$. Now, it should be easy to calculate $d(F^{\flat})$.
You should of course get terms involving $dx^1\wedge dx^2$ etc. Now, write \begin{align} dx^1\wedge dx^2 = \left(\frac{\omega^1}{h_1}\right)\wedge \left(\frac{\omega^2}{h_2}\right) = \frac{1}{h_1h_2}\omega^1\wedge\omega^2 \end{align} And now, since $\{\omega^1,\omega^2,\omega^3\}$ are an orthonormal coframe, the Hodge star is easy to calculate (be careful with signs of course): $\star(\omega^1\wedge \omega^2)=\omega^3$ and so on. Now, you have computed $\star d(F^{\flat})$, and your answer should be in terms of $\omega^1,\omega^2,\omega^3$, so taking the $\sharp$ of this just changes $\omega^i$ to $e_i$. Then you should be done (recall that $f_i= F^ih_i$; no summation). You should notice that the formula can be expressed as a nice determinant as in @K.defaoite's answer.
Then, of course, you can just take the special case of spherical coordinates where you already know the various scale factors $h_r=1$, $h_{\theta}=r, h_{\phi}=r\sin\theta$.