Let $$R = \begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{pmatrix}$$
be a rotation matrix and $A(x) = -x_2 dx_1 + x_1 dx_2$ a differential form, then since $dA = 2 dx_1 \wedge dx_2$ we have $dA = R^* dA.$
Consequently, we have that there is $f$, a $1$-form such that
$$ A - R^*A = df,$$ where $R^*$ denotes the pullback.
However, somehow this computation does not work out for me when I plug in numbers:
I get using vector notation $R^*A(x) =A(Rx) (-\sqrt{3} x/2 + y/2, -x/2 - \sqrt{3} y/2 )$ and thus $$ A(x)-R^*A(x) = \begin{pmatrix} 1/2 (\sqrt{3} x - 3 y)\\ 1/2 (3 x + \sqrt{3} y) \end{pmatrix}$$
but this is not a gradient field
What am I doing wrong?
I believe there is a mistake in your computation. Your transformation is actually a rotation of the plane (in your case, of angle $2\pi/3$): $$ R_\theta=\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}. $$ For any choice of $\theta$, you get $R_\theta^*A=A$:
$$\begin{aligned} R_\theta^*(-ydx+xdy)&=(x\sin\theta+y\cos\theta)d(x\cos\theta-y\sin\theta)+(x\cos\theta-y\sin\theta)d(x\sin\theta+y\cos\theta)\\ &=(\dots)\\ &=-ydx+xdy. \end{aligned}$$ But then $R_\theta^*A-A=0$, and indeed this is of the form $df$ (for some constant function $f$).