We denote the set of nonzero integers and the set of non-negative integers with $\mathbb Z^*$ and $\mathbb N_0$, respectively. Find all functions $f:\mathbb Z^* \to \mathbb N_0$ such that:
$a)$ $f(a+b)\geq min(f(a), f(b))$ for all $a,b$ in $\mathbb Z^*$ for which $a+b$ is in $\mathbb Z^*$.
$b)$ $f(ab)=f(a)+f(b)$ for all $a,b$ in $\mathbb Z^*$.
**
   - The constant function \( f \equiv 0 \) satisfies both conditions:
     \[
     f(a+b) = 0 \geq \min(0, 0) = 0,
     \]
     \[
     f(ab) = 0 = 0 + 0 = f(a) + f(b).
     \]

5. **Non-Trivial Solutions:**
   - Suppose there exists a non-trivial function \( f \). Let \( p \) be the smallest positive integer such that \( f(p) \neq 0 \).
   - If \( p \) is not a prime, then \( p = ab \) for some \( a, b \in \mathbb{Z}^* \):
     \[
     f(p) = f(a) + f(b).
     \]
   - Since \( p \) is the smallest integer with \( f(p) \neq 0 \), both \( f(a) \) and \( f(b) \) must be zero, contradicting the assumption that \( f(p) \neq 0 \). Hence, \( p \) must be a prime.

6. **Behavior for Multiples of \( p \):**
   - For any integer \( k \), \( f(pk) = f(p) + f(k) \). Since \( f(p) > 0 \), \( f(pk) > 0 \).
   - For any \( r \) such that \( 0 < r < p \), \( f(r) = 0 \) because \( p \) is the smallest integer with \( f(p) \neq 0 \).
   - Consider \( f(kp + r) \):
     \[
     f(kp + r) \geq \min(f(kp), f(r)) = \min(f(kp), 0) = 0.
     \]
   - Since \( f(kp + r) \) must be non-negative and \( f(r) = 0 \), it follows that \( f(kp + r) = 0 \).

7. **Conclusion:**
   - The function \( f \) must be of the form \( f(n) = v_p(n) \), where \( v_p(n) \) is the p-adic valuation of \( n \) (the highest power of \( p \) dividing \( n \)).
   - Therefore, the solutions are \( f(x) = 0 \) or \( f(x) = v_p(x) \) for some prime \( p \).

The final answer is \( \boxed{ f(x) = 0 } \) or \( f(x) = v_p(x) \) for some prime \( p \).