jmc

$f(3)=\frac{1+3}{1-3\cdot 3}=-\frac{1}{2}$. Then $f_1(3)=f(-\frac{1}{2})=\frac{1-\frac{1}{2}}{1+3\cdot \frac{1}{2}}=\frac{1}{5}$, $f_2(3)=f(\frac{1}{5})=\frac{1+\frac{1}{5}}{1-3\cdot \frac{1}{5}}=3$ and $f_3(3)=f(3)=\frac{1+3}{1-3\cdot 3}=-\frac{1}{2}$. It follows immediately that the function cycles and $f_n(3)=-\frac{1}{2}$ if $n=3k$, $f_n(3)=\frac{1}{5}$ if $n=3k+1$ and $f_n(3)=3$ if $n=3k+2$. Since $1993=3\cdot 664+1$, $f_{1993}(3)=\boxed{\frac{1}{5}}$.