help

I got 324 but I might be wrong.

If

$f(c)=\dfrac{3}{2c-3}$, find $\dfrac{kn^2}{lm}$

when $f^{-1}(c)\times c \times f(c)$ equals the simplified fraction $\dfrac{kc+l}{mc+n}$,
where $k$, $l$, $m$, and $n$ are integers.

My attempt:

$\begin{array}{|rcll|} \hline f(c) &=& \dfrac{3}{2c-3} \quad | \quad c=f^{-1}(c) \\\\ f\Big(f^{-1}(c)\Big) &=& \dfrac{3}{2f^{-1}(c)-3} \quad | \quad f\Big(f^{-1}(c)\Big) = c \\\\ c &=& \dfrac{3}{2f^{-1}(c)-3} \\\\ c\Big(2f^{-1}(c)-3\Big) &=& 3 \\\\ 2cf^{-1}(c)-3c &=& 3 \\\\ 2cf^{-1}(c) &=& 3c+3 \\\\ cf^{-1}(c) &=& \dfrac{3c+3}{2} \\\\ \mathbf{f^{-1}(c)\times c} &=& \mathbf{\dfrac{3c+3}{2}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline f^{-1}(c)\times c \times f(c) &=& \dfrac{(3c+3)}{2}\times \dfrac{3}{(2c-3)} \\\\ f^{-1}(c)\times c \times f(c) &=& \dfrac{3(3c+3)}{2(2c-3)} \\\\ \mathbf{f^{-1}(c)\times c \times f(c)} &=& \mathbf{\dfrac{9c+9}{4c-6}} \\ \hline && \boxed{k=9,\ l=9,\ m=4,\ n=-6} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{9(-6)^2}{9*4} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{36}{4} \\\\ \mathbf{\dfrac{kn^2}{lm}} &=& \mathbf{9} \\ \hline \end{array}$