I have a 3-manifold with Riemannian metric $g=\omega_1^2+\omega_2^2+\omega_3^2$, where $\omega_i$ are 1-forms (coframe fields).
I am in this situation: $- (u^{-1}p+\frac{3}{2}u^{-1})du \wedge \omega_2 \wedge \omega_3 + \omega_2 \wedge d\omega_3=d\omega_2 \wedge \omega_3$, that it should be possible to write like: $-(u^{-1}p+\frac{3}{2}u^{-1})du \wedge \omega_2 \wedge \omega_3 =d(\omega_2 \wedge \omega_3)$. where $p$ is a constant.
EDITED AFTER "User284193" answer and "jgon" comment:
How I calculate $\omega_2$ and $\omega_3$?
I don't know how calculate $\omega_2$ and $\omega_3$
If I got your question, you can move $\omega_2 \wedge d \omega_3$ on the RHS, so the RHS becomes $d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$.
This is equal to $d(\omega_2 \wedge \omega_3)$, since $d(\omega_2 \wedge \omega_3)=d\omega_2 \wedge \omega_3 + (-1)^1 \omega_2 \wedge d\omega_3=d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$ ($\omega_2$ is a $1$-form)
In general $d(\alpha \wedge \beta)=d \alpha \wedge \beta + (-1)^p \alpha \wedge d \beta$ where $\alpha$ is a $p$-form.