Cauchy integral formula on $f(1/z)$. 
Let $f$ be holomorphic on $\Bbb C\setminus\{0\}$ and $\lim_{|z|\to\infty} f(z) = 0$. Then
$${1\over 2\pi i}\int_{|\zeta| = 1}{f(\zeta)\over \zeta -z}\ d\zeta = -f(z),\quad |z|>1.$$

Since $f$ is bounded on a deleted neighborhood of $\infty$, $\infty$ is an isolated singularity so $f:\Bbb C_{\infty}\setminus\{0\}\to\Bbb C$  is holomorphic. Define a map $g(z) = f(1/z)$ for $z\in\Bbb C$. Then $g$ is entire. By applying CIF to $g$ on $|z|<1$, we get
$$g(z) = {1\over 2\pi i}\int_{|\zeta| =1}{g(\zeta)\over \zeta - z}\ d\zeta.$$
Parametrizing the unit circle $|\zeta| =1$ by $e^{i\theta}$ for $0\leq\theta\leq 2\pi$,
\begin{align*}
{1\over 2\pi i}\int_{|\zeta| =1}{g(\zeta)\over \zeta - z}\ d\zeta & = {1\over 2\pi}\int_0^{2\pi} {g(e^{i\theta})\over e^{i\theta} - z}e^{i\theta}\ d\theta\\
& = {1\over 2\pi}\int_{0}^{2\pi}{f(e^{-i\theta})\over e^{i\theta} - z}e^{i\theta}\ d\theta\\
& = {1\over 2\pi}\int_0^{2\pi}{f(e^{i\theta})\over e^{-i\theta}-z}e^{-i\theta}\ d\theta\\
& = {1\over 2\pi i}\int_{|\zeta| =1}{f(\zeta)\over \zeta - z\zeta^2}\ d\zeta.\\
& = f(1/z).\\
\end{align*}
I think the argument should be like this but something is wrong during the integral manipulation. Could you help?
 A: As you noticed $g(z):=f(1/z)$ $(z \not=0)$, $g(0):=0$ is an entire function. Therefore $h(z):= g(z)/z$ $(z \not=0)$, $h(0):= g'(0)$ is entire, too. Fix any $z$ with $|z|>1$. Then $|1/z| < 1$, thus
$$
zf(z)=h(1/z)= \frac{1}{2\pi i}\int_{|z|=1} \frac{h(\zeta)}{\zeta-1/z} d\zeta =
\frac{1}{2\pi} \int_0^{2\pi} \frac{h(e^{it})e^{it}}{e^{it}-1/z} dt = 
\frac{1}{2\pi} \int_0^{2\pi} \frac{f(e^{-it})e^{-it}}{e^{it}-1/z} e^{it} dt
$$
$$
=\frac{1}{2\pi} \int_0^{2\pi} \frac{f(e^{-it})}{e^{it}-1/z}  dt 
=\frac{1}{2\pi} \int_0^{2\pi} \frac{f(e^{it})}{e^{-it}-1/z}  dt 
=\frac{1}{2\pi} \int_0^{2\pi} \frac{f(e^{it})e^{it}}{1-e^{it}/z}  dt 
$$
$$
=\frac{1}{2\pi} \int_0^{2\pi} \frac{zf(e^{it})e^{it}}{z-e^{it}}  dt
=-\frac{z}{2\pi} \int_0^{2\pi} \frac{f(e^{it})e^{it}}{e^{it}-z}  dt = -\frac{z}{2\pi i}\int_{|z|=1} \frac{f(\zeta)}{\zeta-z} d\zeta
$$
Division by $-z$ yields the desired formula.
A: I think if partial fractions are used instead of the last step, the answer in the question would be correct.  Here is a similar approach.
Since $f(1/\xi)$ is holomorphic for $|\xi|\lt1$
$$
\begin{align}
\frac1{2\pi i}\int_{|\zeta|=1}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta
&=-\frac1{2\pi i}\int_{|\xi|=1}\frac{f(1/\xi)}{1/\xi-z}\,\mathrm{d}(1/\xi)\tag{1a}\\
&=\frac1{2\pi i}\int_{|\xi|=1}\frac{f(1/\xi)}{\xi-z\xi^2}\,\mathrm{d}\xi\tag{1b}\\
&=\frac1{2\pi i}\int_{|\xi|=1}f(1/\xi)\left(\frac1{\xi-0}-\frac1{\xi-1/z}\right)\mathrm{d}\xi\tag{1c}\\[6pt]
&=f(\infty)-f(z)\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a):}$ substitute $\xi=1/\zeta$
$\phantom{\text{(1a):}}$ negate since $1/\zeta$ circles the origin clockwise
$\text{(1b):}$ $\mathrm{d}(1/\xi)=-\frac{\mathrm{d}\xi}{\xi^2}$
$\text{(1c):}$ partial fractions
$\text{(1d):}$ apply Cauchy's Integral Formula to $f(1/\xi)$
$\phantom{\text{(1d):}}$ where $|z|\gt1$ and $f(\infty)=\lim\limits_{|\zeta|\to\infty}f(\zeta)$
