I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that: $$ f\left(\frac{1}{n}\right)=(-1)^n \frac{1}{n^3}$$ for all positive integer $n$.
My idea was to apply the identity theorem for holomorphic funtions. How can I do that? Maybe I must consider the subsequences $ \frac{1}{2k}, \frac{1}{2k+1}$. Can somebody help me?
While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.
Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $\Bbb{C}$, so $f(z)=z^3$ must hold for all $z\in \Bbb{C}$, but then the condition says that $f\big(1/(2k+1)\big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.