Let $\mathbb{D}^2$ be the closed $2$-dimensional unit disk, ebdowed with the Euclidean metric.
Let $\sigma \in \Omega^1(\mathbb{D}^2)$ be a smooth one-form satisfying $d\sigma=\text{const}\cdot dx \wedge dy,\delta \sigma=\text{const}$. ($\delta \sigma=\text{const}$ is equivalent to $d\star \sigma=\text{const} \cdot (dx \wedge dy)$ where $\star$ is the Hodge start operator w.r.t the Euclidean metric.)
In other words, I assume that $\delta \sigma$ is a constant function, and $d\sigma$ is a constant multiple of the standard volume form on $\mathbb{D}^2$.
Is it true that $d\sigma=0,\delta \sigma=0$?
Let $\sigma=f(x,y)dx+g(x,y)dy$ be a generic $1$-form. Then $$d\sigma=-\frac{\partial f}{\partial y}dx\wedge dy + \frac{\partial g}{\partial x}dx\wedge dy$$ and $$\delta \sigma = d*\sigma = d(-g(x,y)dx+f(x,y)dy)=\frac{\partial g}{\partial y}dx\wedge dy + \frac{\partial f}{\partial x}dx\wedge dy$$
Picking for example $f(x,y)=g(x,y)=x$ gives $d\sigma=dx\wedge dy$ and $\delta \sigma= dx\wedge dy$.
In general $2$ differential equations on two functions of two variables still leave you a lot of wiggle room.