Calculating divergence for parabolic coordinates. I'm trying to calculate divergence for parabolic coordinates with a goal to check if i understand all notions of vector analysis correctly.
Definition of parabolic coordinates:
$$x = uv \\ y = \frac{1}{2}(u^2 - v^2)$$
Firstly, I calculate $du$ and $dv$:
$$
dx = v du + udv \\
dy = udu-vdv
$$
$$
dv = \frac{u}{u^2+v^2}dx - \frac{v}{u^2+v^2}dy\\
dv = \frac{v}{u^2+v^2}dx + \frac{u}{u^2+v^2}dy
$$
Then I calculate $e_u$ and $e_v$ - basis vectors for tangent plane in new variables:
$$
x_u = v \\
y_u = u \\
e_u = \frac{(x_u,y_u)}{||(x_u,y_u)||} = (\frac{v}{\sqrt{u^2+v^2}}, \frac{u}{\sqrt{u^2+v^2}})
$$
$$
x_v = u \\
y_v = -v \\
e_v = \frac{(x_v,y_v)}{||(x_v,y_v)||} = (\frac{u}{\sqrt{u^2+v^2}}, -\frac{v}{\sqrt{u^2+v^2}})
$$
Calculating $du(e_u)$ and $dv(e_v)$:
$$
du(e_u) = \frac{u^2+v^2}{\sqrt{u^2+v^2}(u^2+v^2)} = \frac{1}{\sqrt{u^2+v^2}} (=dv(e_v))
$$
Noting that $du(e_v) = dv(e_u) = 0$, let $(F, H)$ be a vector field.
Now I need to calculate area form:
$$dx\wedge dy = (u^2 + v^2)du\wedge dv$$
And take interiour product with my field:
$$(u^2 + v^2)((du)(F,H) dv - dv(F,H) du) \\
= (u^2 + v^2)(\frac{F}{\sqrt{u^2+v^2}}dv - \frac{H}{\sqrt{u^2+v^2}}du) \\
= \sqrt{u^2+v^2}(Fdv-Hdu)$$
Then I apply differential on top of it to get my divergence as coefficient:
$$
d(\sqrt{u^2+v^2}(Fdv-Hdu)) \\
= (\frac{u}{\sqrt{u^2+v^2}}du + \frac{v}{\sqrt{u^2+v^2}}dv)\wedge(Fdv-Hdu) + \sqrt{u^2+v^2}(F_u du\wedge dv - H_v dv \wedge du) \\
= \frac{1}{\sqrt{u^2+v^2}}(uFdu \wedge dv - vH dv \wedge du) + \sqrt{u^2+v^2}(F_u + H_v) du \wedge dv \\
= (\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)) du \wedge dv
$$
So divergence of $(F, H)$ should be of form
$$
\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)
$$
But there I can see that another $\frac{1}{u^2 + v^2}$ was lost somewhere. But I can't find where it was lost, so I'm asking for help to find it.

Thanks for clarifying. OK, so the vector field is dual to the $1$-form $\omega = F\omega_1 + H\omega_2$, where $\omega_1 = \sqrt{u^2+v^2}\,du$ and $\omega_2=\sqrt{u^2+v^2}\,dv$ give an orthonormal coframe. [Note that the metric $dx^2 + dy^2 = \omega_1^2 + \omega_2^2$.]
Of course we have $\star\omega_1 = \omega_2$ and $\star\omega_2=-\omega_1$. So
\begin{align*}
d(\star\omega) &= d(F\omega_2) - d(H\omega_1) = d\big(F\sqrt{u^2+v^2}dv\big) - d\big(H\sqrt{u^2+v^2}du\big)\\
&= \left(\sqrt{u^2+v^2}(F_u+H_v) + F\frac u{\sqrt{u^2+v^2}}+H\frac v{\sqrt{u^2+v^2}}\right)du\wedge dv \\
&= \left(\frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}\right)\omega_1\wedge\omega_2.
\end{align*}
Finally, using $\star(\omega_1\wedge\omega_2) = 1$, we have
$$\text{div} (F,H) = \frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}.$$
