= Solution
Parametrize the positively oriented circle by $z=re^{i\theta}$. The <Cauchy derivative formula> at the origin becomes
$$
f'(0)=\frac1{2\pi r}\int_0^{2\pi}f(re^{i\theta})e^{-i\theta}\,d\theta.
$$
Taking the complex conjugate of the <Cauchy integral theorem> and using $d\bar z=-ir e^{-i\theta}\,d\theta$ gives
$$
0=-ir\int_0^{2\pi}\overline{f(re^{i\theta})}e^{-i\theta}\,d\theta.
$$
Add this zero integral to the derivative integral. Since $f+\bar f=2\operatorname{Re}f=2u$, <recovering an analytic derivative from the boundary real part> yields
$$
\boxed{f'(0)=\frac1{\pi r}\int_0^{2\pi}u(\theta)e^{-i\theta}\,d\theta.}
$$
The <complex conjugation> on the integrand and the reversed sign in $d\bar z$ are both required; the circle's orientation itself has not been reversed.
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