Find all pairs $(f,h)$ of functions $f,h:\mathbb{R}\rightarrow\mathbb{R}$ such as for all real $x$ and $y$ the equation holds. $$f(x^2+yh(x))=xh(x)+f(xy)$$
1. Let \( P(x, y) \) be the assertion \( f(x^2 + y h(x)) = x h(x) + f(xy) \).

2. Let \( a = f(0) \).

3. Consider \( P(x, 0) \):
   \[
   P(x, 0) \implies f(x^2) = x h(x) + f(0) = x h(x) + a
   \]

4. Now consider \( P(x, \frac{x^2}{x - h(x)}) \) for \( x \neq 0 \) and \( h(x) \neq x \):
   \[
   P(x, \frac{x^2}{x - h(x)}) \implies f(x^2 + \frac{x^2 h(x)}{x - h(x)}) = x h(x) + f(\frac{x^3}{x - h(x)})
   \]
   Simplifying the left-hand side:
   \[
   f(x^2 + \frac{x^2 h(x)}{x - h(x)}) = f(\frac{x^2 (x - h(x) + h(x))}{x - h(x)}) = f(\frac{x^3}{x - h(x)})
   \]
   Thus:
   \[
   f(x^2) = x h(x) + f(\frac{x^3}{x - h(x)})
   \]
   Since \( f(x^2) = a \), we have:
   \[
   a = x h(x) + f(\frac{x^3}{x - h(x)})
   \]

5. If \( h(x) = x \):
   \[
   P(x, 0) \implies f(x^2) = x^2 + a
   \]

6. Therefore, for \( x \neq 0 \), either \( f(x^2) = a \) or \( f(x^2) = x^2 + a \).

7. Consider \( P(x, 0) \):
   \[
   f(x^2) = x h(x) + a
   \]
   Thus:
   \[
   \text{Either } a = x h(x) + a \implies h(x) = 0
   \]
   \[
   \text{Or } x^2 + a = x h(x) + a \implies h(x) = x
   \]

8. If \( h(t) = 0 \) for some \( t \neq 0 \):
   \[
   P(t, \frac{x}{t}) \implies f(x) = f(t^2) \text{ (constant)}
   \]
   \[
   P(x, y) \implies h(x) = 0 \text{ for all } x \neq 0
   \]
   Hence, the solution is:
   \[
   \boxed{\text{S1: } f(x) = a \text{ for all } x \text{ and } h(x) = 0 \text{ for all } x \neq 0 \text{ and } h(0) = b}
   \]
   which is indeed a solution for any \( a, b \in \mathbb{R} \).

9. If \( h(0) \neq 0 \):
   \[
   P(0, \frac{x}{h(0)}) \implies f(x) = a \text{ (constant)}
   \]
   Thus, the same solution as above.

10. If \( h(0) = 0 \) and \( h(x) \neq 0 \) for all \( x \neq 0 \):
    \[
    h(x) = x \text{ for all } x
    \]
    \[
    P(x, y) \implies f(x^2 + xy) = x^2 + f(xy)
    \]
    Let \( Q(x, y) \) be the assertion \( f(x^2 + xy) = x^2 + f(xy) \):
    \[
    Q(x, 0) \implies f(x^2) = x^2 + a
    \]
    \[
    Q(x, -x) \implies f(-x^2) = -x^2 + a
    \]
    Thus:
    \[
    f(x) = x + a \text{ for all } x
    \]
    Hence, the solution is:
    \[
    \boxed{\text{S2: } f(x) = x + a \text{ for all } x \text{ and } h(x) = x \text{ for all } x}
    \]
    which is indeed a solution for any \( a \in \mathbb{R} \).