jmc

First, we compute $\frac{1}{r}+\frac{1}{s}+\frac{1}{t}$: We have $$\frac{1}{r}+\frac{1}{s}+\frac{1}{t}=\frac{rs+st+tr}{rst}=\frac{2}{-1}=-2$$by Vieta's formulas. Squaring this equation, we get $${\left(\frac{1}{r}+\frac{1}{s}+\frac{1}{t}\right)}^2=4,$$or $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}+2\left(\frac{1}{rs}+\frac{1}{st}+\frac{1}{tr}\right)=4.$$But we also have $$\frac{1}{rs}+\frac{1}{st}+\frac{1}{tr}=\frac{r+s+t}{rst}=\frac{-9}{-1}=9,$$so $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}+2(9)=4.$$Therefore, $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\boxed{-14}.$$(Note that the left-hand side is a sum of squares, but the right-hand side is negative! This means that some of $r,$ $s,$ and $t$ must be nonreal.)