Continuous $f$ such that $f(x)=f(x^2)$ is constant?

For $x>0$, as $x^{1/2^{n}}\rightarrow 1$, then $f(x)=f(x^{1/2^{n}})\rightarrow f(1)$, so $f(x)=f(1)$. Now $f(0)=\lim_{x\rightarrow 0^{+}}f(x)=f(1)$. And we know that $f$ is an even function.

If $f$ is continous, for all sequence $(x_n)_{n \in \mathbb{N}}$ such as $x_n \underset{n \rightarrow +\infty}{\rightarrow}x$ then $$ f\left(x_n\right) \underset{n \rightarrow +\infty}{\rightarrow}f\left(x\right) $$ It should help you conclude.