If $n = p$ is a prime number then
$$f(p) = f\left(\frac{p}{p}\right) - f(p) = f(1) - f(p)$$
i.e.
$$f(p) = \frac{f(1)}{2} \qquad (1)$$
If $n = pq$ ($p$ and $q$ are prime numbers) then
$$f(n) = f\left(\frac{n}{p}\right) - f(p) = f(q) - f(p) \text{ or } f(n) = f\left(\frac{n}{q}\right) - f(q) = f(p) - f(q)$$
i.e.
$$f(n) = 0 \text{ (because of (1))}.$$
If $n$ is a product of three prime numbers
$$f(n) = f\left(\frac{n}{p}\right) - f(p) = 0 - f(p) = -f(p) = -\frac{f(1)}{2}.$$
By induction on the number of prime divisors, it can be easily shown that if $n$ is a product of $k$ prime numbers, then
$$f(n) = (2-k) \frac{f(1)}{2} \qquad (2)$$
Then, from $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$ and (2) we have
$$\begin{aligned} 2006 &= f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = \\ &= \frac{2-2007}{2}f(1) + \frac{2-2008}{2}f(1) + \frac{2-2009}{2}f(1) = -\frac{3\cdot2006}{2}f(1) \end{aligned}$$
i.e.
$$f(1) = -\frac{2}{3} \qquad (3)$$
Now from $2007 = 3^2 \cdot 223$, $2008 = 2^3 \cdot 251$, $2009 = 7^2 \cdot 41$ and (3) we get
$$\begin{aligned} f(2007^2) + f(2008^3) + f(2009^5) &= \frac{2-6}{2}f(1) + \frac{2-12}{2}f(1) + \frac{2-15}{2}f(1) = \\ &= -\frac{27}{2}f(1) = -\frac{27}{2} \cdot \left(-\frac{2}{3}\right) = 9 \end{aligned}$$