Find all differentiable functions $ f:\mathbb{R}\longrightarrow\mathbb{R} $ that verify the conditions:
$ \text{(i)}\quad\forall x\in\mathbb{Z} \quad f'(x) =0 $

$ \text{(ii)}\quad\forall x\in\mathbb{R}\quad f'(x)=0\implies f(x)=0 $ 
1. **Given Conditions:**
   - \( f:\mathbb{R} \to \mathbb{R} \) is differentiable.
   - \( \forall x \in \mathbb{Z}, \quad f'(x) = 0 \).
   - \( \forall x \in \mathbb{R}, \quad f'(x) = 0 \implies f(x) = 0 \).

2. **Claim:**
   The function \( f(x) = 0 \) for all \( x \in [n, n+1] \) for all \( n \in \mathbb{N} \).

3. **Proof:**
   - From the first condition, \( f'(n) = 0 \) for all \( n \in \mathbb{Z} \).
   - By the second condition, \( f'(n) = 0 \implies f(n) = 0 \). Therefore, \( f(n) = 0 \) for all \( n \in \mathbb{Z} \).

4. **Interval Analysis:**
   - Consider the interval \([n, n+1]\) for any \( n \in \mathbb{Z} \).
   - Since \( f(n) = 0 \) and \( f(n+1) = 0 \), and \( f \) is continuous and differentiable, we need to analyze the behavior of \( f \) within this interval.

5. **Application of Fermat's Theorem:**
   - Fermat's Theorem states that if \( f \) has a local extremum at \( c \) in the open interval \((n, n+1)\), then \( f'(c) = 0 \).
   - If \( f \) is not constant in \([n, n+1]\), it must attain a local extremum at some point \( c \in (n, n+1) \).

6. **Contradiction:**
   - Suppose \( f \) is not constant in \([n, n+1]\). Then there exists \( c \in (n, n+1) \) such that \( f'(c) = 0 \).
   - By the second condition, \( f'(c) = 0 \implies f(c) = 0 \).
   - Therefore, \( f \) must be zero at all points where its derivative is zero, including all local extrema.

7. **Conclusion:**
   - Since \( f \) is zero at all local extrema and at the endpoints \( n \) and \( n+1 \), \( f \) must be zero throughout the interval \([n, n+1]\).
   - This argument holds for all \( n \in \mathbb{Z} \).

8. **Final Conclusion:**
   - Since \( f \) is zero on every interval \([n, n+1]\) for all \( n \in \mathbb{Z} \), and these intervals cover the entire real line, \( f(x) = 0 \) for all \( x \in \mathbb{R} \).

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = 0 } \) for all \( x \in \mathbb{R} \).