Is it correct? $\operatorname{Res}(f(z),\infty) = - \operatorname{Res}(f(1/z),0)$ If define $g(w) = f(1/w)$, can we say that $f(z)$'s residue at $\infty$ can be derived from $g(w)$'s residue at $0$? i.e.
$$\operatorname{Res}(f(z),\infty) = - \operatorname{Res}(g(w),0)$$
 A: No.  If you do not know about "meromorphic differentials" ...
Take the Laurent series
$$
f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n
$$
If this series is valid in a punctured neighborhood of $0$, then $\text{Res}(f(z),z=0) = a_{-1}$.  If this series is valid in a punctured neighforhood of $\infty$, then $\text{Res}(f(z),z=\infty) = -a_{-1}$.
Why the minus sign?  One way to think of this is: if a closed curve surrounds $0$ once counterclockwise, then it surrounds $\infty$ once clockwise.
Consider $f(1/z)$.  Its Laurent series is
$$
f\left(\frac{1}{z}\right) = \sum_{n=-\infty}^{+\infty} a_n z^{-n} = 
\sum_{n=-\infty}^{+\infty} a_{-n} z^{n}.
$$
So
$$
\text{Res}(f(1/z),z=0) = a_1
$$
and (in most cases) $a_1 \ne -a_{-1}$.  The equation in the OP is wrong.

What is the right formula?
We can consider $-f(1/z)/z^2$.  Then
$$
-\frac{1}{z^2}\;f\left(\frac{1}{z}\right)  =
 -\sum_{n=-\infty}^{+\infty} a_{-n} z^{n-2} =
 \sum_{n=-\infty}^{+\infty} (-a_{-n-2}) z^{n}
$$
and then
$$
\text{Res}(-f(1/z)/z^2,z=0) = -a_{-1} = \text{Res}(f(z),z=\infty) .
$$
