I have an ellipsoid given by $E\subset\mathbb{R}^{3}$ such that $x^{2}+y^{2}+2z^{2}=1$ with smooth immersion $f:E\to\mathbb{R}^{3}$. I am asked to compute $\mathrm{d}(\omega\wedge\ast\omega)$ where $\omega:=f^{\ast}\mathrm{d}x$.
My attempt so far:
$$\mathrm{d}(\omega\wedge\ast\omega)=\mathrm{d}\omega\wedge\ast\omega-\omega\wedge \mathrm{d}(\ast\omega)=\underbrace{\mathrm{d}(f^{\ast}\mathrm{d}x)}_{=f^{\ast}d^{2}x=0}\wedge\ast(f^{\ast}\mathrm{d}x)-(f^{\ast}\mathrm{d}x)\wedge \mathrm{d}(\ast(f^{\ast}\mathrm{d}x))\\=-(f^{\ast}\mathrm{d}x)\wedge \mathrm{d}(\ast(f^{\ast}\mathrm{d}x))$$
But I am not sure how to simplify this expression further. Alternatively, I think I could maybe also use the identity
$$(\omega\wedge\ast\omega)=\langle\omega,\omega\rangle\operatorname{vol}_{g}$$
where $g$ is the induced metric on $E$, but I am not sure if this helps.
HINT: You need coordinates on your ellipsoid. Try elliptical (stretched) spherical coordinates \begin{align*} x &= \sin u\cos v \\ y &= \sin u\sin v \\ z &= 2\cos u. \end{align*} Now pull back the Euclidean metric and do the computations in $(u,v)$-coordinates. You should find that the pullback of the metric is $(1+3\sin^2u)\,du^2 + \sin^2u\, dv^2$, so $\omega_1 = \sqrt{1+3\sin^2u}\,du$ and $\omega_2 = \sin u\,dv$ give an orthonormal coframe. This means that $\star\omega_1 = \omega_2$ and $\star\omega_2 = -\omega_1$.