Solution
= Solution
At any $z\neq0$, take sufficiently small $h$ so $z+h\neq0$. The complex difference quotient is
$$
\frac{(z+h)^{-1}-z^{-1}}h=-\frac1{z(z+h)}\longrightarrow-\frac1{z^2}.
$$
Hence the complex <derivative> exists at every point of the punctured plane, and
$$
\boxed{f'(z)=-z^{-2},\qquad f\text{ is holomorphic on }\mathbb C\setminus\{0\}.}
$$
The excluded point is essential to the domain statement: the reciprocal has a <simple pole> there.