If the function $f$ is defined by \[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac{3}{2}$, then $c$ is
$\text{(A)} \ -3 \qquad  \text{(B)} \ - \frac{3}{2} \qquad  \text{(C)} \ \frac{3}{2} \qquad  \text{(D)} \ 3 \qquad  \text{(E)} \ \text{not uniquely determined}$
As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$
. However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that $c=-3 \Rightarrow \boxed{A}$.