= Solution
Use the <Wirtinger derivatives> $\partial_z=(\partial_x-i\partial_y)/2$ and $\partial_{\bar z}=(\partial_x+i\partial_y)/2$, and area measure $dA=dx\,dy=(i/2)dz\wedge d\bar z$. The supplied boundary-integral identity is the planar <Generalized Stokes theorem>; the usual <Poincare lemma> is a different local exactness result.
Let $f\in C^1(\overline D)$ and $z\in D$. Remove a disk of radius $\epsilon$ around $z$, and apply <Generalized Stokes theorem> to the <one-form> $f(\zeta)d\zeta/(\zeta-z)$ on the punctured domain. Away from the puncture,
$$
d\left(\frac{f(\zeta)}{\zeta-z}d\zeta\right)
=\frac{f_{\bar\zeta}(\zeta)}{\zeta-z}\,d\bar\zeta\wedge d\zeta
=2i\frac{f_{\bar\zeta}(\zeta)}{\zeta-z}\,dA(\zeta).
$$
The outer boundary is counterclockwise and the small inner circle clockwise. Its counterclockwise integral tends to $2\pi i f(z)$. Passing to the limit gives the <Cauchy-Pompeiu formula>
$$
\boxed{f(z)=\frac1{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}\,d\zeta
-\frac1\pi\int_D\frac{f_{\bar\zeta}(\zeta)}{\zeta-z}\,dA(\zeta).}
$$
The weak $1/|\zeta-z|$ singularity is locally integrable. If the boundary term vanishes on expanding $D$ to the whole plane, the formula becomes $f(z)=\pi^{-1}\int_{\mathbb C}f_{\bar\zeta}(\zeta)/(z-\zeta)\,dA(\zeta)$. In particular it yields the distributional normalization $\partial_{\bar z}[1/(\pi z)]=\delta_0$. For <holomorphic functions> the area term vanishes and one recovers the <Cauchy integral formula>.
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