Answer: $f(n) = n^3 - 1, n \in N_0$.

From the given conditions we observe that: $f(0) = f(1) = 0$. For $n \ge 2$, define $g(n) = f(n) + 1$. Then
$g(2) = 8$ and
$$\begin{aligned} g(mn) &= f(mn) + 1 = f(m) + f(n) + f(m)f(n) + 1 \\ &= (f(m) + 1)(f(n) + 1) \\ &= g(m)g(n), \text{ for all } m, n \ge 2. \end{aligned}$$

Fix an integer $n > 2$, and consider a sequence of rational numbers $\{p_k/q_k, k \ge 1\}$, each term of which is greater than $\log_2 n$ and which converges to $\log_2 n$. Then from $n < 2^{p_k/q_k}$ we get $n^{q_k} < 2^{p_k}$, and by the monotonicity of $g$ we obtain
$$g(n^{q_k}) \le g(2^{p_k}).$$
By the multiplicativity of $g$ we get
$$g(n) \ge g(2)^{p_k/q_k} = 2^{3p_k/q_k} = (2^{p_k/q_k})^3.$$
Letting $k \to \infty$, then $g(n) \le n^3$. Arguing similarly, we get $g(n) \ge n^3$. Hence $g(n) = n^3$. Therefore $f(n) = n^3 - 1, \forall n$ is the unique solution satisfying the given conditions.