Given $f$ is an entire function and for any $z\neq0$, $f$ satisfies $f(z)=f(\frac{1}{z^2})$.
The question asks to prove that $f$ is constant.
My approach: For any $|z|>1$, we have $\dfrac{1}{|z|}<1$ hence by Maximum Modulus Principle, $|f(z)|\leq|f(1)|$ for any $z\in\mathbb{C}$.
But then I got stuck at the next step... Is my direction correct? If yes, what should be the next step? If no, how can we prove $f$ is constant?
By continuity,we have $$\lim_{z\to \infty}f(z)=\lim_{z\to \infty}f\left(\dfrac{1}{z^2}\right)=f(0)$$ If $\lim_{z\to \infty}f(z)\to \infty$, then $f(0)$ also tends to infinity which contradicts the $f$ is entire.
Thus $f$ is uniformly bounded, and by Liouville's theorem, f is constant.