Solution:

Let $a, b, c$ be the zeros of $f$. Then $f(x) = (x - a)(x - b)(x - c)$. Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$, so $g(x) = k(x - a^{2})(x - b^{2})(x - c^{2})$ for some constant $k$. Since $a b c = -f(0) = -1$, we have $k = k a^{2} b^{2} c^{2} = -g(0) = 1$. Thus,
$$g(x^{2}) = (x^{2} - a^{2})(x^{2} - b^{2})(x^{2} - c^{2}) = (x - a)(x - b)(x - c)(x + a)(x + b)(x + c) = -f(x) f(-x)$$
Setting $x = 3$ gives $g(9) = -f(3) f(-3) = -(31)(-29) = 899$.