Denote by $\mathbb{Z}^{*}$ the set of all nonzero integers and denote by $\mathbb{N}_{0}$ the set of all nonnegative integers. Find all functions $f:\mathbb{Z}^{*} \rightarrow \mathbb{N}_{0}$ such that:

$(1)$ For all $a,b \in \mathbb{Z}^{*}$ such that $a+b \in \mathbb{Z}^{*}$ we have $f(a+b) \geq $ [b]min[/b] $\left \{ f(a),f(b) \right \}$.

$(2)$ For all $a, b \in \mathbb{Z}^{*}$ we have $f(ab) = f(a)+f(b)$.
1. **Claim**: Any function \( f \) that satisfies the given conditions must be of the form \( f(n) = k \cdot v_p(n) \) for some prime \( p \) and integer \( k \geq 0 \), where \( v_p(n) \) is the largest \( \alpha \in \mathbb{N}_0 \) such that \( p^\alpha \mid n \).

2. **Verification**: We need to verify that any such \( f \) satisfies the conditions:
   - For all \( a, b \in \mathbb{Z}^* \) such that \( a + b \in \mathbb{Z}^* \), we have \( f(a + b) \geq \min \{ f(a), f(b) \} \).
   - For all \( a, b \in \mathbb{Z}^* \), we have \( f(ab) = f(a) + f(b) \).

3. **Initial Conditions**:
   - Taking \( a = b = 1 \), we find \( f(1) = 2f(1) \), yielding \( f(1) = 0 \).
   - Taking \( a = b = -1 \), we find \( f(1) = 2f(-1) \), which implies \( f(-1) = 0 \). Hence, \( f(-a) = f(a) \) for all \( a \).

4. **Restriction to Nonnegative Integers**: Since \( f(-a) = f(a) \), we can restrict our attention to \( \mathbb{N} \).

5. **Prime Powers**:
   - For every prime \( p \) and positive integer \( k \), we have \( f(p^k) = kf(p) \).
   - For \( n = \prod_{1 \leq j \leq n} p_j^{\alpha_j} \), we obtain \( f(n) = \sum_{1 \leq j \leq n} \alpha_j \varphi(p_j) \), where \( \varphi \) is the restriction of \( f \) onto the primes.

6. **Non-zero \( \varphi \)**:
   - If \( \varphi = 0 \) everywhere, we are done.
   - Assume there is a prime \( p \) such that \( \varphi(p) \neq 0 \).

7. **Contradiction for Multiple Primes**:
   - Assume \( \varphi(\cdot) \) is positive on two distinct primes \( p \) and \( q \).
   - Consider \( a = pq - pq^p \) and \( b = pq^p \). Then \( a + b = pq \), forcing \( f(a + b) = f(p) + f(q) \).
   - Notice \( f(b) = f(p) + pf(q) \) and 
     \[
     f(a) = f\left(pq \cdot (1 - q^{p-1})\right) = f(p) + f(q) + f\left(q^{p-1} - 1\right) \geq 2f(p) + f(q).
     \]
     Using Fermat's theorem, \( q^{p-1} - 1 \equiv 0 \pmod{p} \), hence \( f\left(q^{p-1} - 1\right) \geq f(p) \).

8. **Contradiction**:
   - We have \( f(p) + f(q) = f(a + b) \) and 
     \[
     \min \{ f(a), f(b) \} \geq \min \{ f(p) + pf(q), 2f(p) + f(q) \}.
     \]
   - This leads to a contradiction, hence \( \varphi(\cdot) \) is non-zero on at most one prime.

9. **Conclusion**: The function \( f \) must be of the form \( f(n) = k \cdot v_p(n) \) for some prime \( p \) and integer \( k \geq 0 \).

The final answer is \( \boxed{ f(n) = k \cdot v_p(n) } \) for some prime \( p \) and integer \( k \geq 0 \).