jmc

Let $r,$ $s,$ and $t$ be the roots of $f(x),$ so that $f(x)=(x-r)(x-s)(x-t)$. Then $r^2,$ $s^2,$ and $t^2$ are the roots of $g,$ so we can write $$g(x)=A(x-r^2)(x-s^2)(x-t^2)$$for some constant $A.$ Taking $x=0,$ we get $$-1=-A{r}^2{s}^2{t}^2.$$We know that $rst=-1$ by Vieta, so $$-1=-A(-1{)}^2=-A$$and $A=1.$ Then $$g(x)=(x-r^2)(x-s^2)(x-t^2),$$so $$g(9)=(9-r^2)(9-s^2)(9-t^2).$$To evaluate this product, we write $$\begin{array}{rl}g(9)& =(3-r)(3+r)(3-s)(3+s)(3-t)(3+t)\\ & =(3-r)(3-s)(3-t)(3+r)(3+s)(3+t)\\ & =(3-r)(3-s)(3-t)[-(-3-r)(-3-s)(-3-t)]\\ & =f(3)\cdot -f(-3).\end{array}$$We know that $f(x)=(x-r)(x-s)(x-t),$ so in particular, $31=f(3)=(3-r)(3-s)(3-t)$ and $-29=f(-3)=(-3-r)(-3-s)(-3-t).$ Therefore, $$g(9)=f(3)\cdot -f(-3)=31\cdot 29=\boxed{899}.$$