I'm doing this exercise, but I don't know if my solution is good.
Show that can't exists any entire function $f$ such that $f(\frac{1}{n})=|\frac{1}{n}|^3$, $ n \in \mathbb{Z} \setminus 0$.
I started writing $t=\frac{1}{n}$. So $f(t)=|t|^3$. Since $0$ is an accumulation point for {$z \in \mathbb{C}: f(z)=0$}, then $f$ must be $f=0$. But that's not possible, since $f(t)$ is different from zero for every $t$ in its domain.
If such a function $f$ existed, let $g(z)=z^3$. Then$$(\forall n\in\mathbb{N}):f\left(\frac1n\right)=g\left(\frac1n\right).$$Therefore, by the identity theorem, $f=g$. But $f(-1)=1$, whereas $g(-1)=-1$.
One problem with your approach lies in the fact that you tried to apply the identity theorem to a non-analytic function.