If $n=p$ is a prime number, we have

$$f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)$$

i.e.

$$f(p)=\frac{f(1)}{2}$$

If $n=p q$, where $p$ and $q$ are prime numbers, then

$$f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0$$

If $n$ is a product of three prime numbers, we have

$$f(n)=f\left(\frac{n}{p}\right)-f(p)=0-f(p)=-f(p)=-\frac{f(1)}{2}$$

With mathematical induction by a number of prime multipliers we shall prove that: if $n$ is a product of $k$ prime numbers then

$$f(n)=(2-k) \frac{f(1)}{2}$$

For $k=1$, clearly the statement (2) holds.

Let statement (2) hold for all integers $n$, where $n$ is a product of $k$ prime numbers.

Now let $n$ be a product of $k+1$ prime numbers. Then we have $n=n_{1} p$, where $n_{1}$ is a product of $k$ prime numbers.

So

$$f(n)=f\left(\frac{n}{p}\right)-f(p)=f\left(n_{1}\right)-f(p)=(2-k) \frac{f(1)}{2}-\frac{f(1)}{2}=(2-(k+1)) \frac{f(1)}{2}$$

So (2) holds for every integer $n>1$.

Now from $f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006$ and because of (2) we have

$$\begin{aligned} 2006 & =f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right) \\ & =\frac{2-2007}{2} f(1)+\frac{2-2008}{2} f(1)+\frac{2-2009}{2} f(1)=-\frac{3 \cdot 2006}{2} f(1) \end{aligned}$$

i.e.

$$f(1)=-\frac{2}{3}$$

Since

$$2007=3^{2} \cdot 223,2008=2^{3} \cdot 251,2009=7^{2} \cdot 41$$

and because of (2) and (3), we get

$$\begin{aligned} f\left(2007^{2}\right)+f\left(2008^{3}\right)+f\left(2009^{5}\right) & =\frac{2-6}{2} f(1)+\frac{2-12}{2} f(1)+\frac{2-15}{2} f(1) \\ & =-\frac{27}{2} f(1)=-\frac{27}{2} \cdot\left(-\frac{2}{3}\right)=9 \end{aligned}$$