Find all functions $f: R \to R$ such that $f (xy) \le yf (x) + f (y)$, for all $x, y\in R$.
1. **Initial Claim**: We claim that \( f = 0 \) is the only solution. To verify this, we start by analyzing the given functional inequality:
   \[
   f(xy) \le y f(x) + f(y) \quad \forall x, y \in \mathbb{R}
   \]

2. **Substitution**: Let us substitute \( y = 1 \) into the inequality:
   \[
   f(x \cdot 1) \le 1 \cdot f(x) + f(1) \implies f(x) \le f(x) + f(1)
   \]
   This inequality is always true and does not provide new information. However, it does not contradict our claim.

3. **Non-negativity**: Next, we substitute \( y = -1 \):
   \[
   f(x \cdot (-1)) \le (-1) f(x) + f(-1) \implies f(-x) \le -f(x) + f(-1)
   \]
   Let \( x = 1 \):
   \[
   f(-1) \le -f(1) + f(-1) \implies 0 \le -f(1)
   \]
   This implies \( f(1) \ge 0 \).

4. **Boundedness**: Now, consider \( y = -z \):
   \[
   f(x(-z)) \le (-z) f(x) + f(-z) \implies f(-xz) \le -z f(x) + f(-z)
   \]
   If \( f \) were unbounded, we could make \( f(x) \) arbitrarily large, which would contradict the inequality since \( z \) is fixed. Therefore, \( f \) must be bounded.

5. **Upper Bound**: Suppose there exists a constant \( c \) such that \( f(x) \le c \) for all \( x \). Then:
   \[
   f(xy) \le y f(x) + f(y) \le y c + c = c(y + 1)
   \]
   Setting \( y = -1 \):
   \[
   f(-x) \le -f(x) + f(-1)
   \]
   Since \( f \) is bounded, let \( c \) be the upper bound. Then:
   \[
   0 \le f(-x) \le -f(x) + c
   \]
   This implies \( f(x) \le c \) and \( f(-x) \le c \).

6. **Zero Function**: To show \( f(x) = 0 \), assume \( f(x) \neq 0 \) for some \( x \). Then:
   \[
   f(xy) \le y f(x) + f(y)
   \]
   If \( f(x) > 0 \), let \( y = -1 \):
   \[
   f(-x) \le -f(x) + f(-1)
   \]
   Since \( f \) is bounded and non-negative, \( f(x) \) must be zero to satisfy the inequality for all \( x \). Therefore, \( f(x) = 0 \) for all \( x \).

\(\blacksquare\)

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