A 1 - Form That is Not The Differential of a 0 - Form

If
$\omega^1$
is a 1 - form in
$\mathbb{R}^3$
then we can write
$\omega^1 = f_1 dx_1 +f_2 dx_2 +f_3 dx_3$

If
$omega^0$
is a 0 - form in
$\mathbb{R}^3$
then we can write
$\omega^0 =g(x_1,x_2,x_3) \rightarrow d \omega^0 = \frac{\partial g}{\partial x_1} dx_1 + \frac{\partial g}{\partial x_2} dx_2 + \frac{\partial g}{\partial x_3} dx_3$

Equating
$\omega^1 , \: d \omega^0$
gives
$f_1 = \frac{\partial g}{\partial x_1} , \: f_2 = \frac{\partial g}{\partial x_2} , \; f_3 = \frac{\partial g}{\partial x_3}$
or with
$\mathbf{f} =(f_1 ,f_2,f_3)^T , \: \mathbf{f} = \mathbf{\nabla} g$
.
The existence of a scalar function
$g$
such that for a vector field
$\mathbf{f}$
we have
$\mathbf{f} = \mathbf{\nabla} g$
is one of the properties of a conservative vector field, or equivalently
$\mathbf{\nabla} \times \mathbf{f} =0$

Take
$f_1 =x_2 , \: f_2=-x_1 \: f_3=0$
then
$(\mathbf{\nabla} \times \mathbf{f} = (\frac{\partial}{\partial x_1} , \frac{\partial}{\partial x_2} , \frac{\partial}{\partial x_3} )^T \times (x_2, -x_1,0)^T = -2$

Hence no such function
$g$
exists.