Computing differential of a $0$-form

Question

I am stuck on problem 30.1 from Munkres' Analysis on Manifolds. The question is to directly compute that $d(d\omega)=0$, given that $\omega = xy dx + 3 dy - yz dz$. So far I found that this $\omega$ we're dealing with is a 0-form, so theorem 30.3 in Munkres gives us that $d\omega$ would be defined by $(D_{1}f dx_{1}+D_{2}f dx_{2}+D_{3}f dx_{3})$. But I don't really understand how to use this theorem. What would be our $f$ in this case? And do we use $dx, dy, dz$ as $dx_1,dx_2,dx_3$, respectively? While not really understanding what $f$ stands for in this formula, I have computed the following now: $d\omega = D_{1}f dx+D_{2}f dy+D_{3}f dz=xdx-ydz$, but this does not seem to get me anywhere.

Answer 1

As stated in the comments, $\omega$ is a one-form. The general formula for the exterior derivative of a one-form $\omega=\sum_i f_i\ dx^i$ is $$d\omega=\sum_{i,j} \frac{\partial f_i}{\partial x^j}dx^j\wedge dx^i.$$ If we compute that for the given $\omega$, we get $$d\omega=y\ dx\wedge dx+x\ dy\wedge dx+0-z\ dy\wedge dz-y\ dz\wedge dz=-x\ dx\wedge dy -z\ dy\wedge dz.$$ Similarly, if we compute $d(d\omega),$ we get $$d(d\omega)=-dx\wedge dx\wedge dy-dz\wedge dy\wedge dz=0.$$