Here is the problem.
A nonconstant rational function over the real numbers(a rational function is a function that can be expressed as $\frac{p(x)}{q(x)}$, with $p(x), q(x)$ as polynomial function) $f(x)$, is defined such that:
$(f(x))^2 - a = f(x^2)$ for all $x$ ( $a$ is a constant value)
Prove that $f(x)$ must be of the form $x^k$ for some constant $k$.
I have an idea of how to prove it, but I can't do so rigorously. I can prove that any polynomial with more than $2$ terms, or monomials, would not work. But I can't prove it won't work for any rational function.
If $q(z)$ is nonconstant, then $q(z^2)$ has at least one complex zero that $q(z)$ doesn't have. (Choose, for example, the zero $z_1$ of $q(z^2)$ with minimal nonzero argument.) This means that $q(z)$ must be constant, for otherwise the right-hand side is undefined at a point where the left-hand side is defined. This reduces to the case where $f$ is a polynomial, which you have (almost?) solved.
Edit: wnoise points out that this argument fails if $q(z)$ is a power of $z$. Perhaps the argument for polynomials extends to Laurent polynomials (where negative powers of $z$ are allowed)?