North Macedonia

If $n=p$ is a prime number then $$f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)$$ i.e. $$f(p)=\frac{f(1)}{2}(1)$$ 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 $$f(n)=(2-k)\frac{f(1)}{2}(2)$$ Then, from $f(2^{2007})+f(3^{2008})+f(5^{2009})=2006$ and (2) we have $$\begin{array}{rl}2006& =f(2^{2007})+f(3^{2008})+f(5^{2009})=\\ & =\frac{2-2007}{2}f(1)+\frac{2-2008}{2}f(1)+\frac{2-2009}{2}f(1)=-\frac{3\cdot 2006}{2}f(1)\end{array}$$ i.e. $$f(1)=-\frac{2}{3}(3)$$ Now from $2007=3^2\cdot 223$, $2008=2^3\cdot 251$, $2009=7^2\cdot 41$ and (3) we get $$\begin{array}{rl}f(2007^2)+f(2008^3)+f(2009^5)& =\frac{2-6}{2}f(1)+\frac{2-12}{2}f(1)+\frac{2-15}{2}f(1)=\\ & =-\frac{27}{2}f(1)=-\frac{27}{2}\cdot \left(-\frac{2}{3}\right)=9\end{array}$$