Show Cauchy-Pompeiu type formula Let $\mathbb{D}$ be the open unit disc, $\overline{\mathbb{D}}$ its closure, and $f$ a $C^{1}$ function on a neighborhood of $\mathbb{D}$. Show that for all $a\in \mathbb{D}$
\begin{equation*}
f(a)= -\frac{1}{\pi} \int_{\overline{\mathbb{D}}} \frac{\partial f}{\partial \overline{z}}  \frac{1-|z^{2}|}{1-\overline{z}a} \frac{1}{z-a} d\lambda(z) + \frac{1}{\pi}\int_{\overline{\mathbb{D}}} \frac{f(z)}{(1-\overline{z}a)^{2}} d\lambda(z)
\end{equation*}
with $\lambda$ being the Lebesgue measure.
In my opinion this is most likely an application of Stokes formula, with the integrand being the exterior derivative of a well-chosen differential form. But I do not know which form...
 A: I'm going to answer my own question in case someone is interested.
Consider the differential form $\displaystyle\omega=f(z)\frac{1-|z|^{2}}{(z-a)(1-\bar{z}a)}dz$, as the integral strongly suggests.
We apply Stokes' theorem for $K_{\varepsilon}=\mathbb{D}\setminus B(a,\varepsilon)$ (with $\varepsilon$ small enough):
\begin{equation*}
\int_{\partial K_{\varepsilon}}\omega = \int_{K_{\varepsilon}}d\omega .
\end{equation*}
Since $|z|=1$ on $\partial \mathbb{D}$, 
\begin{equation*}
\int_{\partial \mathbb{D}} \omega =0.
\end{equation*}
With the parametrisation $z=a+\varepsilon e^{i\theta}$, 
\begin{equation*}\int_{\partial B(a,\varepsilon)} \omega=i\int_{0}^{2\pi} \frac{1-|a+\varepsilon e^{i\theta}|^{2}}{1-\bar{a}(a+\varepsilon e^{i\theta})} f(a+\varepsilon e^{i\theta}).
\end{equation*}
By continuity of $f$, 
\begin{equation*}
\int_{\partial B(a,\varepsilon)} \omega \underset{\varepsilon\rightarrow 0}{\longrightarrow}  2i\pi f(a)
\end{equation*}
hence
\begin{equation*}
\int_{\partial K_{\varepsilon}}\omega \underset{\varepsilon\rightarrow 0}{\longrightarrow}  -2i\pi f(a).
\end{equation*}
Let us compute $d\omega$.
\begin{equation*}
d\omega= \frac{\partial}{\partial\bar{z}}\left(f(z)\frac{1-|z|^{2}}{(z-a)(1-\bar{z}a)}\right) d\bar{z} \wedge dz = 2i\frac{\partial}{\partial\bar{z}}\left(f(z)\frac{1-|z|^{2}}{(z-a)(1-\bar{z}a)}\right) dx \wedge dy .
\end{equation*}
Some computations using the fact that $\displaystyle\frac{\partial}{\partial\bar{z}}(uv)=\frac{\partial}{\partial\bar{z}}(u)v+ \frac{\partial}{\partial\bar{z}}(u)v$ and $\displaystyle\frac{\partial u}{\partial\bar{z}} = \overline{\frac{\partial \bar{u}}{\partial z}}$ show that:
\begin{equation*}
dw=2i\left(\frac{\partial f}{\partial\bar{z}}\frac{1-|z|^{2}}{(z-a)(1-\bar{z}a)} - \frac{f(z)}{(1-\bar{z}a)^{2}}\right) dx\wedge dy
\end{equation*}
Since $z\mapsto 1/(z-a)$ is locally integrable and $z \mapsto 1/(1-\bar{z}a)$ has no poles in $\mathbb{D}$, Lebesgue Dominated Convergence Theorem shows that
\begin{equation*}
\int_{K_{\varepsilon}}d\omega\underset{\varepsilon\rightarrow 0}{\longrightarrow} \int_{\mathbb{D}}d\omega.
\end{equation*}
Putting all of the previous results together we get the desired equality.
