Polynomials of polynomials

Since we are dealing with polynomials, let's try to find the degree. Define $d = \deg(f)$

$\deg(f(f(x) + f(-x))) = \deg(f(f(x))) = d^2$

$\deg(9f(x^2)) = \deg(f(x^2)) = 2d$

For this equation to be true for all real $x$ , we must have $d^2 = 2d \Rightarrow d(d - 2) = 0$ and thus,

$\deg(f) = 0, 2$

We are dealing with either a constant or a quadratic. Let us assume the latter, but keep in mind that $f$ could be constant and so the leading coefficient could be $0$ . We can now define $f(x) = ax^2 + bx + c$ , and we will start with the LHS (left-hand side):

$f((ax^2 + bx + c) + (ax^2 - bx + c)) = f(2ax^2 + 2c)$

$= a(2ax^2 + 2c)^2 + b(2ax^2 + 2c) + c$

$= (4a^3)x^4 + (8a^2c + 2ab)x^2 + (4ac^2 + 2bc + c)$

And the RHS (right-hand side):

$9(a(x^2)^2 + b(x^2) + c) = (9a)x^4 + (9b)x^2 + (9c)$

If both sides are to be equal over all $x$ , the coefficients must be equal, so we get:

$4a^3 = 9a \Rightarrow a(4a^2 - 9) = 0 \Rightarrow a(2a - 3)(2a + 3) = 0 \Rightarrow a = 0, \pm\frac{3}{2}$

$8a^2c + 2ab = 9b$

$4ac^2 + 2bc + c = 9c$

So we have $(a, b, c) = (0, 0, 0), (\pm\frac{3}{2}, 0, 0)$ as the only possible coefficients. The first one $(0, 0, 0)$ is just a constant $f(x) = 0$ , and so we now have the only three polynomials that satisfy the original equation. The maximum value of $f(9)$ is from the coefficients $(\frac{3}{2}, 0, 0)$ giving:

$\max(f(9)) = \frac{3}{2}9^2 = \frac{3*81}{2} = \frac{243}{2}$

So the answer is $243*2 = \boxed{486}$