We claim that we must have $f(x)=x^{3}-3 x$. First, note that the condition $f(x)+f(-x)=0$ implies that $f$ is odd. Combined with $f$ being monic, we know that $f(x)=x^{3}+a x$ for some real number $a$. Note that $a$ must be negative; otherwise $f(x)$ and $f(f(x))$ would both be increasing and 1 would be the only possible value of $g(y)$. Now, consider the condition that the set of possible values of $g(y)$ is $\{1,5,9\}$. The fact that we can have $g(y)=9$ means that some horizontal line crosses the graph of $f(f(x)) 9$ times. Since $f(f(x))$ has degree 9, this means that its graph will have 4 local maxima and 4 local minima. Now, suppose we start at some value of $y$ such that $g(y)=9$, and slowly increase $y$. At some point, the value of $g(y)$ will decrease. This happens when $y$ is equal to a local maximum of $f$. Since $g(y)$ must jump from 9 down to 5, all four local maxima must have the same value. Similarly, all four local minima must also have the same value. Since $f$ is odd, it suffices to just consider the four local maxima. The local maximum of $f(x)$ occurs when $3 x^{2}+a=0$. For convenience, let $a=-3 b^{2}$, so $f(x)=x^{3}-3 b^{2} x$. Then, the local maximum is at $x=-b$, and has a value of $f(-b)=2 b^{3}$. We consider the local maxima of $f(f(x))$ next. They occur either when $x=-b$ (meaning $f(x)$ is at a local maximum) or $f(x)=-b$. If $f(x)=-b$, then $f(f(x))=f(-b)=2 b^{3}$. Thus, we must have $f(f(-b))=f\left(2 b^{3}\right)=2 b^{3}$. This yields the equation $$f\left(2 b^{3}\right)=8 b^{9}-3 b^{2} \cdot 2 b^{3}=2 b^{3}$$ which factors as $2 b^{3}\left(b^{2}-1\right)\left(2 b^{2}+1\right)^{2}$. The only possible value of $b^{2}$ is 1. Thus, $f(x)=x^{3}-3 x$, and our answer is $10^{3}-3 \cdot 10=970$.