I have $dx=x^{1/2}\omega_1$, (so $\omega_1=x^{-1/2}dx$), and $\star dx=x^{1/2}\omega_2 \wedge \omega_3$ (where $\star$ is the Star Hodge Operator).
Then I have $d(x^{1/2}\omega_2 \wedge \omega_3)= -p \omega_1 \wedge \omega_2 \wedge \omega_3$, where $p$ is a constant.
What will $\omega_2 \wedge \omega_3$ be?
Doing the calculation, $$ \begin{aligned} \mathrm{d}\bigl(x^{p+1/2}\,\omega_2\wedge\omega_3\bigr) &= \mathrm{d}\bigl(x^p\,\,x^{1/2}\,\omega_2\wedge\omega_3\bigr) = p x^{p-1}\mathrm{d}x\wedge x^{1/2}\,\omega_2\wedge\omega_3+ x^p\wedge\mathrm{d}\bigl(x^{1/2}\,\omega_2\wedge\omega_3\bigr)\\ &= p x^{p-1}\,x^{1/2}\omega_1\wedge x^{1/2}\,\omega_2\wedge\omega_3 + x^p\wedge\bigl(-p\omega_1\wedge\omega_2\wedge\omega_3\bigr)\\ &= (p-p) \,x^{p}\omega_1\wedge\omega_2\wedge\omega_3 = 0, \end{aligned} $$ one sees that $x^{p+1/2}\,\omega_2\wedge\omega_3$ is a closed $2$-form. Thus, there always exist (localy) functions $y$ and $z$ such that $$ x^{p+1/2}\,\omega_2\wedge\omega_3 = \mathrm{d}y\wedge\mathrm{d}z, $$ which is about all that one can say. Namely $$ \omega_2\wedge\omega_3 = x^{-(p+1/2)}\,\mathrm{d}y\wedge\mathrm{d}z $$ locally for some functions $y$ and $z$, where $\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\not=0$.
