Let $α = dz - ydx \in Ω^1 (\mathbb{R}^3)$
Find a vector field Z in $\mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$
What I did:
I computed $d\alpha = dx \wedge dy $. Then let $Z=a\partial_x + b\partial_y +c\partial_z $ be a vector field in $\mathbb{R}^3$
We have $α(Z) = 1 \Rightarrow c\partial_z - ya\partial_x = 1 $ (I)
and $dα(Z,.) = 0 \Rightarrow dx \wedge dy (a\partial_x + b\partial_y +c\partial_z\ ,\ .)$.
but we have $ dx \wedge dy (a\partial_x) = dx(a \partial_x)dy(.) -dx(.)dy(a \partial_x) = ady(.)$
and $ dx \wedge dy (b\partial_y) = dx(b \partial_y)dy(.) -dx(.)dy(b \partial_y) = -bdx(.)$
and $ dx \wedge dy (b\partial_y) = 0$
which means finally $ ady(.) = bdx(.) $ (II)
Could we find a,b and c from the two results (I) and (II)?
Thank you!
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=\partial_z$ which is readily seen to be correct.