Let $F(x)=\sum_{i=1}^nx_i^4$ on $\mathbb R^n$, and let $\Omega=\{x\in\mathbb R^n:F(x)\leq 1\}$. Consider $$ \omega=\sum_{i=1}^n(-1)^{i+1}D_iF\,dx_1\wedge\dots\hat{dx_i}\dots\wedge dx_n. $$ a) Show that $\int_{\partial\Omega}\omega>0$.
b) Let $g=(g_1,\dots,g_n)\colon\Omega\to\partial\Omega$ be a map of class $C^\infty$. Show that $$ \sum_{i=1}^n D_iF(g)dg_i=0. $$
I managed to show part a using Stokes' theorem. However, I'm don't know how to solve b.
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After some help here and there, I finally found a solution. I've posted it as an answer, as there is really nothing left to ask.
We know that $d(F\circ g)=g^*(dF)$. The exterior derivative of $F$ is given by $$ dF=\sum_{i=1}^n D_iF\,dx_i. $$ So we're going to calculate the pull-back: $$ g^*(dF)=\sum_{i=1}^ng^*(D_iF\,dx_i)=\sum_{i=1}^n g^*(D_iF)g^*(dx_i)=\sum_{i=1}^nD_iF(g)\,dg_i, $$ where we used that $g^*dx_i=dg_i$. Now I need to argue that this equals zero. We know that $F(x)\equiv1$ on its boundary, and since $g(\Omega)=\partial\Omega$, we have $F\circ g\equiv 1$. This means that $d(F\circ g)=0$. This concludes the prove.