\[\omega^1\]
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.