Problem. If a function $f$ is $\mathbb{R}$-differentiable in a neighborhood of a point $a \in \mathbb{C}$ then $$ \lim _{\varepsilon \rightarrow 0} \frac{1}{\varepsilon^2} \int_{\{|z-a|=\varepsilon\}} f(z) d z=2 \pi i \frac{\partial f}{\partial \bar{z}}(a) . $$ It's an exercise on "Introduction to complex analysis; volume 1" by B. V. Shabat Chapter 2 and the book gives the following :
Hint: use the formula
$$
f(z)=f(a)+\frac{\partial f}{\partial z}(a)(z-a)+\frac{\partial f}{\partial \bar{z}}(a)(\bar{z}-\bar{a})+o(|z-a|)
$$ and the example
Example: Let $\gamma: I \rightarrow \mathbb{C}$ be an arbitrary piecewise smooth path and $n \neq 1$. We also assume that the path $\gamma(t)$ does not pass through the point $z=0$ in the case $n<0$. The chain rule implies that $\frac{d}{d t} \gamma^{n+1}(t)=(n+1) \gamma^n(t) \gamma^{\prime}(t)$ so that
$$
\int_\gamma z^n d z=\int_\alpha^\beta \gamma^n(t) \gamma^{\prime}(t) d t=\frac{1}{n+1}\left[\gamma^{n+1}(\beta)-\gamma^{n+1}(\alpha)\right]
$$
I guess maybe use $$2\pi i = \int_{\{|z-a|=\varepsilon\}} \frac{dz}{z-a}.$$ But I still don't understand ; please give more hint.