jmc

The cubic equation can be rewritten as $x^3-5x^2+6x-9=0$. First, we list the relations that Vieta's formulas give us: $$\begin{array}{rl}-(r+s+t)& =-5,(♣)\\ rs+rt+st& =6,(♢)\\ -rst& =-9.(♡)\end{array}$$We wish to compute $$\frac{rs}{t}+\frac{rt}{s}+\frac{st}{r}=\frac{r^2{s}^2+r^2{t}^2+s^2{t}^2}{rst}.$$The denominator is $rst=9$. To obtain the numerator, we square equation $(♢)$ to get $$r^2{s}^2+r^2{t}^2+s^2{t}^2+2r^{2}st+2r{s}^{2}t+2rs{t}^2=36.$$We can rewrite this as $$r^2{s}^2+r^2{t}^2+s^2{t}^2+2rst(r+s+t)=36.$$From equations $(♣)$ and $(♡)$, we have $$2rst(r+s+t)=2\cdot 9\cdot 5=90,$$so $$r^2{s}^2+r^2{t}^2+s^2{t}^2=36-90=-54.$$Finally, we have $$\frac{rs}{t}+\frac{rt}{s}+\frac{st}{r}=\frac{r^2{s}^2+r^2{t}^2+s^2{t}^2}{rst}=\frac{-54}{9}=\boxed{-6}.$$