Find all functions $f,g,h: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x) - g(y) = (x-y) \cdot h(x+y)$ for $x,y \in \mathbb{R}.$
1. Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   f(x) - g(y) = (x - y) \cdot h(x + y)
   \]

2. Consider \( P(x, 0) \):
   \[
   f(x) - g(0) = x \cdot h(x)
   \]
   Let \( g(0) = c \). Then:
   \[
   f(x) = x \cdot h(x) + c
   \]

3. Consider \( P(0, y) \):
   \[
   f(0) - g(y) = -y \cdot h(y)
   \]
   Let \( f(0) = k \). Then:
   \[
   k - g(y) = -y \cdot h(y) \implies g(y) = k + y \cdot h(y)
   \]

4. Substitute \( f(x) = x \cdot h(x) + c \) and \( g(y) = k + y \cdot h(y) \) into the original equation:
   \[
   (x \cdot h(x) + c) - (k + y \cdot h(y)) = (x - y) \cdot h(x + y)
   \]
   Simplify:
   \[
   x \cdot h(x) - y \cdot h(y) + c - k = (x - y) \cdot h(x + y)
   \]
   Let \( c = k \):
   \[
   x \cdot h(x) - y \cdot h(y) = (x - y) \cdot h(x + y)
   \]

5. Consider \( P(x, -x) \):
   \[
   f(x) - g(-x) = (x + x) \cdot h(0)
   \]
   \[
   x \cdot h(x) + c - (k - x \cdot h(-x)) = 2x \cdot h(0)
   \]
   Since \( c = k \):
   \[
   x \cdot h(x) + x \cdot h(-x) = 2x \cdot h(0)
   \]
   \[
   h(x) + h(-x) = 2h(0)
   \]

6. Substitute \( y = -y \) in the original equation:
   \[
   x \cdot h(x) + y \cdot h(-y) = (x + y) \cdot h(x - y)
   \]
   Using \( h(-y) = 2h(0) - h(y) \):
   \[
   x \cdot h(x) + y \cdot (2h(0) - h(y)) = (x + y) \cdot h(x - y)
   \]
   \[
   x \cdot h(x) + 2yh(0) - y \cdot h(y) = (x + y) \cdot h(x - y)
   \]

7. Substitute \( y = x - 1 \):
   \[
   x \cdot h(x) + 2(x - 1)h(0) - (x - 1) \cdot h(x - 1) = (x + (x - 1)) \cdot h(x - (x - 1))
   \]
   \[
   x \cdot h(x) + 2(x - 1)h(0) - (x - 1) \cdot h(x - 1) = (2x - 1) \cdot h(1)
   \]
   This implies:
   \[
   h(x) = dx \quad \text{for some} \quad d \in \mathbb{R}
   \]

8. Therefore, we have:
   \[
   f(x) = dx^2 + c \quad \text{and} \quad g(x) = dx^2 + c
   \]

The final answer is \( \boxed{ f(x) = g(x) = dx^2 + c } \) for some \( d \in \mathbb{R} \).