I have to show that $\omega=-4xy\:\mathrm{d}x\wedge \mathrm{d}y-2xz\:\mathrm{d}z\wedge \mathrm{d}x +2yz\:\mathrm{d}y\wedge \mathrm{d}z$ is exact finding a primitve of $\omega$ (by Poincare lemma I already it is).
Now I'm not sure how can I proceed to solve this. This much I know:
For a 1-form $\eta$ given by $\eta =a\:\mathrm{d}x+b\:\mathrm{d}y+c\:\mathrm{d}z$ I have $$\mathrm{d}\eta = \left(\frac{\partial b}{\partial x} -\frac{\partial a}{\partial y}\right)\mathrm{d}x\wedge \mathrm{d}y+\left(\frac{\partial a}{\partial z}\mathrm{d}z - \frac{\partial c}{\partial x}\right)\mathrm{d}z\wedge \mathrm{d}x+\left(\frac{\partial c}{\partial y} - \frac{\partial b}{\partial z}\right)\mathrm{d}y\wedge \mathrm{d}x$$
Thus
$$\frac{\partial b}{\partial x} - \frac{\partial a}{\partial y}=-4xy$$ $$\frac{\partial a}{\partial z} - \frac{\partial c}{\partial x} = -2xz$$ $$\frac{\partial c}{\partial y} - \frac{\partial b}{\partial z} =2yz$$
One of the former exercises I solved let me put $a=0$ or $c=0$ and then find a solution for the remaining variables. Now if I do that, let $a=0$, then this means that
$$\frac{\partial b}{\partial x} = -4xy \implies b = -2x^2y + C_1(y,z)$$ $$\frac{\partial c}{\partial x} = 2xz \implies c=x^2z +C_2(y,z)$$
Now it seems that a solution could be $C_1(y,z)=0$ and $C_2(y,z)=y^2z$ because then $\frac{\partial c}{\partial y} - \frac{\partial b}{\partial z} =2yz$.
Is this correct?.
Now, I'd like to find a general solution for this, how could I do that?. Is this similar to a 1-variable differential equation in the sense that the general solution could be written as the sum of a particular solution and an exact form?; but if that is the case, how can I find the exact form that gives me a general solution?
There is a explicit integral formula. See for example the proof of Poincare Lemma at Calculus in Varieties (Spivak) or my answer in Poincaré lemma for star shaped domain.