Suppose that $f:\mathbb{C} \to \mathbb{C}$ is such a function. Then $|f(z)|>0$ and hence $1/|f(z)|< \infty$. Since $f$ is bounded, by Liouville's theorem it is some constant $C$. By that I conclude than $f(z) = 1/C$ which is constant?
Is this correct?
It is not correct, as the counterexample $f(z)=e^z$ shows (already noted in the comments). The fact that $|f(z)|>0$ does not exclude that $|f(z)|$ comes arbitrarily close to $0$, making $\frac{1}{|f(z)|}$ unbounded. Your argument works if there is a constant $c>0$ such that $$ |f(z)|\ge c.$$