Here are some hints:
Note that for $z \neq 0$, $f(z) = {1 \over z^2} - 2 {1 \over z^4} + \sum_{n=0}^\infty (-1)^n {z^n \over (n+2)!}$.
Let $r(z) = \sum_{n=0}^\infty (-1)^n {z^n \over (n+2)!}$, note that $r$
is entire, and we can write $r(z) = \sum_{n=0}^\infty {r^{(n)}(1) \over n!} (z-1)^n$, with
$r^{(n)}(1) = \sum_{n=0}^\infty (-1)^n n (n-1)\cdots (n-k+1){1 \over (n+2)!}$
For $|z-1| <1$ we can write
${1 \over z^k} = ( 1+(z-1))^{-k} = \sum_{n=0}^\infty \binom{-k}{n}(z-1)^n$.
For $|z-1| >1$ we can write
${1 \over z^k} = {1 \over (1+(z-1))^{k} } = {1 \over (z-1)^k} {1 \over (1+{1 \over (z-1)) } )^{k} } = {1 \over (z-1)^k} \sum_{n=0}^\infty \binom{-k}{n}{1 \over (z-1)^n } $