Let $\omega = z^2 dx \land dy \land dw + yw^3 dy \land dz \land dw + z w dx \land dy \land dz$
We then have $d \omega = 2zdx \land dy \land dw + (w^3 dy + 3yw^2 dw) dy \land dz \land dw + (wdz + z dw) dx \land dy \land dz$.
I'm not sure I know how to expand the brackets, how can I proceed from here? And if the derivative is zero, how can I check if the form is exact?
The basic rules of exterior derivative are those: $$ d(df) = 0,\\ d(f\wedge g) = df\wedge g + (-1)^p f\wedge dg, $$ where $f$ is $p$-form.
For local coordinates it means that: $$ df=d(h(x,y)du\wedge dv) = h'_x(x,y)dx\wedge du\wedge dv + h'_y(x,y)dy\wedge du\wedge dv. $$
In your example, you differentiate $z^2$, but lose $dz$ part.
The expansion of brackets is very easy actually, since exterior product is distributive: $$ (\alpha+\beta)\wedge \omega = \alpha\wedge\omega + \beta\wedge\omega. $$
Finally, from general understanding, you differentiate 3-form of 4 coordinates, so you should get a 4-form. There is only one 4-form in 4 coordinates up to the coefficient: $$h(x,y,z,w)dx\wedge dy\wedge dz \wedge dw.$$ So you should what kind of answer you are going to get.