$f(n)$ is a power of a prime. If $f(n)$ is a power of an odd prime then $f(f(n))=2$ If $f(n)$ is a power of $2$ then $f(f(f(n)))=f(f(2^a))=f(3)=2$

If $f(n)=2$, there is nothing to prove. Let's suppose $f(n)\neq 2$, then $f(n)$ is an eve number and we can represent it as $f(n)=2^t\cdot k$, where $t\geq 1$ and $k\geq 1$ are integers, with $k$ odd. By the definition of $f$ and taking into account that $\frac{f(n)}{2}\mid n$, we have that $n=2^{t-1}\cdot m$, where $w$ is an odd number, with $k\mid m$. If $k\geq 3$, then $2^{t}<f(n)$ and $2^{t}\nmid n$, contradicting the minimality of $f(n)$. Therefore $k=1$, $f(n)=2^t$ and $f(f(n))=f(2^t)=3$, so we conclude that $f(f(f(n)))=f(3)=2$.