jmc

Apply the definition of $f$ to the identity $f(f^{-1}(c))=c$ to find $$\begin{array}{rl}c& =\frac{3}{2f^{-1}(c)-3}\Rightarrow \\ c(2f^{-1}(c)-3)& =3\Rightarrow \\ 2f^{-1}(c)-3& =\frac{3}{c}\Rightarrow \\ 2f^{-1}(c)& =\frac{3}{c}+3\Rightarrow \\ f^{-1}(c)& =\frac{3}{2c}+\frac{3}{2}\Rightarrow \\ & =\frac{3}{2}\left(\frac{1}{c}+1\right).\end{array}$$Therefore, $f^{-1}(c)\times c\times f(c)$ can be found: $$\begin{array}{rl}{f}^{-1}(c)\times c\times f(c)& =\left(\frac{3}{2}\left(\frac{1}{c}+1\right)\right)\times c\times \frac{3}{2c-3}\Rightarrow \\ & =\frac{3}{2}\times \frac{1+c}{c}\times c\times \frac{3}{2c-3}\Rightarrow \\ & =\frac{3\times (1+c)\times 3}{2\times (2c-3)}\Rightarrow \\ & =\frac{9+9c}{4c-6}\Rightarrow \\ & =\frac{9c+9}{4c-6}.\end{array}$$Thus, $k=9$, $l=9$, $m=4$, and $n=-6$. So, $\frac{k{n}^2}{lm}=\frac{9\times (-6{)}^2}{9\times 4}=\boxed{9}$.