2016 is absolutely awesome 5

The important fact to realize here is that $x = -3$ is a fixed point for $f$ .

In other words, $f(-3) = -3$ , and so by induction $f_n(-3) = -3$ .

Now, by the chain rule, notice that:

$f_n^{\,'}(x)=[f(f_{n-1}(x))]^{'}=f^{\,'}(f_{n-1}(x))f_{n-1}^{\,'}(x)$ .

$f_n^{\,'}(-3)=f^{\,'}(f_{n-1}(-3))f_{n-1}^{'}(-3)=f^{\,'}(-3)f_{n-1}^{\,'}(-3)$ .

In other words, to get the derivative of the next $f_n$ at $x = -3$ , we simply multiply the derivative of the previous $f_{n-1}$ at $x = -1$ by the same constant: $f^{\,'}(-3)$ .

So we really only need to compute the very first $f^{\,'}(-3)$ , and then the rest of the derivatives will follow easy from the recursive formula.

Since $f^{\,'}(x)=\dfrac{-18}{(3+2x)^2}$ and hence $f^{\,'}(-3)=-2$ .

We have $f_n^{\,'}(-3)=(-2)^n$ and in particular $f_{2016}^{\,'}(-3)=(-2)^{2016}=2^{2016}$ .

Since $2^{2016}\equiv 536 \pmod{1000}$ so the answer is $\boxed{536}$ .

$\begin{aligned} f_1(x) & = \frac{9}{3+2x} \\ \Rightarrow f_1' (x) & = \frac{-9(2)}{(3+2x)^2} = -\frac{2}{9} (f_1(x))^2 \\ f_2' (x) & = \frac{-18f_1'(x)}{(3+2f_1(x))^2} = -\frac{2}{9} (f_2(x))^2 f_1'(x) = \left(-\frac{2}{9} \right)^2 (f_2(x))^2 (f_1(x))^2 \\ f_3' (x) & = \left(-\frac{2}{9} \right)^3 (f_3(x))^2 (f_2(x))^2 (f_1(x))^2 \\ ... & ... \\ \Rightarrow f_n' (x) & = \left(-\frac{2}{9} \right)^n \prod_{k=1}^n (f_k(x))^2 \end{aligned}$

Now, we note that: $f_1(-3) = \dfrac{9}{3+2(-3)} = -3 \quad \Rightarrow f_2(-3) = \dfrac{9}{3+2f_1(-3)} = -3 \quad \Rightarrow f_3(-3) = \dfrac{9}{3+2f_2(-3)} = -3 \\ \Rightarrow f_n(-3) = -3$

$\begin{aligned} \Rightarrow f_n' (-3) & = \left(-\frac{2}{9} \right)^n (-3)^{2n} = 2^n \\ f_{2016}' (-3) & = 2^{2016} \equiv \boxed{536} \pmod {1000} \end{aligned}$