Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f(x+y)\leq f(x^2+y)$$ for all $x,y$.
To find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that 
\[ f(x+y) \leq f(x^2 + y) \]
for all \( x, y \in \mathbb{R} \), we proceed as follows:

1. **Consider the case \( y = 0 \):**
   \[ f(x) \leq f(x^2) \]
   This inequality must hold for all \( x \in \mathbb{R} \).

2. **Consider the case \( y = -x \):**
   \[ f(0) \leq f(x^2 - x) \]
   This inequality must hold for all \( x \in \mathbb{R} \).

3. **Consider the case \( y = 1 - x \):**
   \[ f(1) \leq f(x^2 + 1 - x) \]
   This inequality must hold for all \( x \in \mathbb{R} \).

4. **Consider the case \( y = -x^2 \):**
   \[ f(-x^2) \leq f(0) \]
   This inequality must hold for all \( x \in \mathbb{R} \).

5. **Consider the case \( y = x^2 - x \):**
   \[ f(x^2 - x) \leq f(x^4 - x^2 + x) \]
   This inequality must hold for all \( x \in \mathbb{R} \).

6. **Let \( 0 \leq d \leq \frac{1}{4} \):**
   We can find \( x_1 \) such that \( x_1^2 - x_1 = d \) and \( x_2 \) such that \( x_2^2 - x_2 = -d \).

7. **Using the periodicity argument:**
   It follows that 
   \[ f(x + d) \leq f(x) \leq f(x + d) \]
   for all \( x \). Hence, 
   \[ f(x) = f(x + d) \]
   for all \( x \). So \( f \) is periodic with each period in \(\left[0, \frac{1}{4}\right]\) and hence constant.

8. **Conclusion:**
   Since \( f \) is periodic with any period in \(\left[0, \frac{1}{4}\right]\), it must be constant. Let \( f(x) = c \) for some constant \( c \).

The final answer is \( \boxed{ f(x) = c } \) for some constant \( c \).