jmc

We have $f_1(x)=\frac{2(3x+1)-9}{3(3x+1)}=\frac{6x-7}{9x+3}.$ We compute the first few $f_n,$ hoping to see a pattern: $$\begin{array}{rl}{f}_2(x)& =f_1\left(\frac{6x-7}{9x+3}\right)=\frac{6\cdot \frac{6x-7}{9x+3}-7}{9\cdot \frac{6x-7}{9x+3}+3}=\frac{6(6x-7)-7(9x+3)}{9(6x-7)+3(9x+3)}=\frac{-27x-63}{81x-54}=\frac{-3x-7}{9x-6},\\ f_3(x)& =f_1\left(\frac{-3x-7}{9x-6}\right)=\frac{6\cdot \frac{-3x-7}{9x-6}-7}{9\cdot \frac{-3x-7}{9x-6}+3}=\frac{6(-3x-7)-7(9x-6)}{9(-3x-7)+3(9x-6)}=\frac{-81x}{-81}=x.\end{array}$$Since $f_3(x)=x$ for all $x,$ we see that $f_k(x)=f_{k-3}(x)$ for all $x.$ Since $1001\equiv 2(\mathrm{mod}3),$ we have $$f_{1001}(x)=f_2(x)=\frac{-3x-7}{9x-6}=x-3,$$so $$\begin{array}{rl}-3x-7& =9x^2-33x+18\\ 0& =9x^2-30x+25=(3x-5{)}^2.\end{array}$$Thus, $x=\boxed{\frac{5}{3}}.$