If we define the operator $$\partial_{\bar{z}} = \frac{\partial_x + i \partial_y}{2}$$ then it is claimed that for $\lambda \in \mathbb{R}$ we have the following series of equalities:
$$ \frac{1}{\pi} \int_{\mathbb{R}^2} \frac{\partial_{\bar{z}} f(x+ i y) }{\lambda - x- y} dx dy = \frac{1}{2 \pi i} \int_{\mathbb{R}^2} \partial_{\bar{z}} \left( \frac{f(x + i y)}{\lambda - x- iy} \right) dz d\bar{z} = \frac{1}{2 \pi i} \int_{\mathbb{R}^2} d \left( \frac{f(x + i y)}{\lambda - x- iy} dz \right) $$
where the operator $d$ is the differential in the sense of a $~1$-form.
I'm having trouble proving both equalities and do not have any idea what the statement "is the differential in the sense of a $~1$-form" means in this context.
The first equality they claim follows from the observation $\partial_{\bar{z}} (\lambda - z)^{-1}= 0$ which I can prove.
You need to work with differential forms and the wedge product (which is anti-commutative) for this to make sense (and be correct). Note that $$dz\wedge d\bar z = d(x+iy)\wedge d(x-iy) = (dx + i\,dy)\wedge (dx -i\,dy) = -2i dx\wedge dy,$$ so $\dfrac{dz\wedge d\bar z}{-2i} = dx\wedge dy$. So they have a sign error in the equation. They mean to write—or perhaps you mistyped?—$d\bar z\wedge dz$, and then it's correct. And then we have $d(g(z)\,dz) = \partial_{\bar z} g \,d\bar z\wedge dz$, which gives the last equality. Here we use the fact that $$dg = \dfrac{\partial g}{\partial z}dz + \dfrac{\partial g}{\partial\bar z}d\bar z$$ and apply $dz\wedge dz = 0$ to get their result.