Find all functions $f:\mathbb{R}\to\mathbb{R}$ for which 
\[ x(f(x+1)-f(x)) = f(x), \] 
for all $x\in\mathbb{R}$ and 
\[ | f(x) - f(y) | \leq |x-y| , \] 
for all $x,y\in\mathbb{R}$. 

[i]Mihai Piticari[/i]
1. **Analyzing the first functional equation:**
   \[
   x(f(x+1) - f(x)) = f(x)
   \]
   For \( x \neq 0 \), we can rewrite this as:
   \[
   f(x+1) - f(x) = \frac{f(x)}{x}
   \]
   This suggests that the difference between \( f(x+1) \) and \( f(x) \) is proportional to \( \frac{f(x)}{x} \).

2. **Considering the second condition:**
   \[
   |f(x) - f(y)| \leq |x - y|
   \]
   This implies that \( f \) is a Lipschitz continuous function with Lipschitz constant 1.

3. **Exploring the implications of the first equation:**
   Let's assume \( f(x) = xg(x) \) for some function \( g(x) \). Substituting \( f(x) = xg(x) \) into the first equation, we get:
   \[
   x((x+1)g(x+1) - xg(x)) = xg(x)
   \]
   Simplifying, we obtain:
   \[
   (x+1)g(x+1) - xg(x) = g(x)
   \]
   \[
   (x+1)g(x+1) = g(x) + xg(x)
   \]
   \[
   g(x+1) = \frac{g(x) + xg(x)}{x+1}
   \]
   \[
   g(x+1) = g(x)
   \]
   This implies that \( g(x) \) is a constant function. Let \( g(x) = c \). Therefore, \( f(x) = cx \).

4. **Verifying the second condition with \( f(x) = cx \):**
   \[
   |f(x) - f(y)| = |cx - cy| = |c||x - y|
   \]
   For the inequality \( |f(x) - f(y)| \leq |x - y| \) to hold, we must have \( |c| \leq 1 \).

5. **Conclusion:**
   The function \( f(x) = cx \) satisfies both the given conditions if and only if \( |c| \leq 1 \).

The final answer is \( \boxed{ f(x) = cx } \) where \( |c| \leq 1 \).