Taking $m=n=1$ we deduce $f(1)=0$. Now, taking $m=1$, we obtain $f(n^2)=f(n)(1+\alpha)$. We now inspect $f(N)$ for $N=p^2q^2$. Taking $m=p^2$ and $n=q$ we obtain $f(N) = f(mn^2) = f(mn)+\alpha f(n) = f(p^2 q)+\alpha f(q)$. Next, we inspect $f(p^2q)$. Note that taking $m=q$ and $n=p$, we have $f(p^2q) = f(mn^2)=f(mn)+\alpha f(n) = f(pq)+\alpha f(p)$. Consequently, $f(p^2 q^2)=f(pq)+\alpha f(p)+\alpha f(q)$. Notice, on the other hand, that $f(p^2q^2)=(1+\alpha)f(pq)$. These two together yield $f(pq)=f(p)+f(q)$ for every $p,q\in\mathbb{N}$. In particular, taking $p=q$ we find that $\alpha=1$. Now, we claim $f$ is constant on primes. Indeed, let $q_1,q_2$ be two distinct primes. Find that $f(q_1q_2)=f(q_1)+f(q_2)$. Using now ${\rm (ii)}$ we find that $f(q_1)\mid f(q_2)$, $f(q_2)\mid f(q_1)$; and $f(q_1),f(q_2)>0$. This implies $f(q_1)=f(q_2)$. Let $f(q)=c$ for $q$ prime. From here, we recover that if $n=\prod_{1\le i\le L}p_i^{\alpha_i}$ where $p_1<\cdots<p_L$ are distinct primes, then \[ f(n) = c \sum_{1\le i\le L}\alpha_i. \]In the case $n=1$ (that is $\alpha_i=0$ for every $i$), we indeed find that $f(n)=0$. This concludes the solution.