$w$ is a form, find $\alpha$ s.t $d \alpha=w$
$w=(2y-4)dy \wedge dz+(y^2-2x)dz \wedge dx+(3-x-2yz)dx \wedge dy$
$w$ is a 2-form and $d \alpha=w$ so $\alpha$ is a 1-form s.t:
$\alpha =Mdx+Ndy+Pdz$
$d \alpha =(\dfrac{\partial P}{dy}-\dfrac{\partial N}{dz} )dy \wedge dz+(\dfrac{\partial M}{dz}-\dfrac{\partial P}{dx} )dz \wedge dx+(\dfrac{\partial N}{dx}-\dfrac{\partial M}{dy} )dx \wedge dy$
And then we equate the respective parts:
$\dfrac{\partial P}{dy}-\dfrac{\partial N}{dz} =2y-4$
$\dfrac{\partial M}{dz}-\dfrac{\partial P}{dx}=y^2-2x$
$\dfrac{\partial N}{dx}-\dfrac{\partial M}{dy} =3-x-2yz$
But I'm not sure how to proceed from here to find M,n, and P I can't find a systematic way to solve this.
As always thank you for your time!
The first test $dω=0$ passes.
Since $d(α+df)=dα$, one can arrange that, for instance, $P=0$. Then $$ -\frac{∂N}{∂z}=2y−4\quad\text{and}\quad \frac{∂M}{∂z}=y^2−2x $$ so that $N=2yz-4z+G(x,y)$ and $M=y^2z-2xz+F(x,y)$. Inserting these into the third rotation component gives $$ 3−x−2yz=\frac{∂N}{∂x}−\frac{∂M}{∂y}=\frac{∂G}{∂x}-2yz-\frac{∂F}{∂y} $$ which can be satisfied in different ways, e.g. by $(F,G)=(xy,3x)$ so that \begin{align} α &=(y^2z-2xz+xy)dx + (2yz-4z+3x)dy \\[0.5em] &=(y^2z+xy)dx+(3x-4z)dy+(x^2-y^2)dz + d(y^2z-x^2z) \end{align}