jmc

From the condition $f^{-1}(x)=f(x),$ $f(f^{-1}(x))=f(f(x)),$ which simplifies to $f(f(x))=x.$

Note that $$\begin{array}{rl}f(f(x))& =f\left(\frac{cx}{2x+3}\right)\\ & =\frac{c\cdot \frac{cx}{2x+3}}{2\cdot \frac{cx}{2x+3}+3}\\ & =\frac{c^{2}x}{2cx+3(2x+3)}\\ & =\frac{c^{2}x}{(2c+6)x+9}.\end{array}$$Setting this equal to $x,$ we get $$\frac{c^{2}x}{(2c+6)x+9}=x,$$so $c^{2}x=(2c+6)x^2+9x.$ We want this to hold for all $x,$ so we require the corresponding coefficients on both sides to be equal. In other words, from the quadratic term we get $0=2c+6$, and from the linear terms we get $c^2=9$. This gives us $c=\boxed{-3}.$