Find all functions $f : R \to R$ which satisfy $f(x) = max_{y\in R} (2xy - f(y))$ for all $x \in  R$.
1. We start with the given functional equation:
   \[
   f(x) = \max_{y \in \mathbb{R}} (2xy - f(y))
   \]
   This implies that for all \( x \in \mathbb{R} \), \( f(x) \) is at least as large as any value of \( 2xy - f(y) \) for any \( y \in \mathbb{R} \). Therefore, we have:
   \[
   f(x) \geq 2xy - f(y) \quad \forall x, y \in \mathbb{R}
   \]

2. Setting \( y = x \) in the inequality \( f(x) \geq 2xy - f(y) \), we get:
   \[
   f(x) \geq 2x^2 - f(x)
   \]
   Simplifying this, we obtain:
   \[
   2f(x) \geq 2x^2 \implies f(x) \geq x^2 \quad \forall x \in \mathbb{R}
   \]

3. Next, we need to show that there exists some \( y \in \mathbb{R} \) such that:
   \[
   f(x) = 2xy - f(y)
   \]
   Rearranging this equation, we get:
   \[
   f(x) + f(y) = 2xy
   \]

4. To solve this, we assume \( f(x) = x^2 \) and check if it satisfies the given functional equation. Substituting \( f(x) = x^2 \) into the equation \( f(x) + f(y) = 2xy \), we get:
   \[
   x^2 + y^2 = 2xy
   \]
   Rearranging, we have:
   \[
   (x - y)^2 = 0 \implies x = y
   \]
   This shows that \( f(x) = x^2 \) is a potential solution.

5. Finally, we verify that \( f(x) = x^2 \) satisfies the original functional equation:
   \[
   f(x) = \max_{y \in \mathbb{R}} (2xy - f(y))
   \]
   Substituting \( f(y) = y^2 \), we get:
   \[
   \max_{y \in \mathbb{R}} (2xy - y^2)
   \]
   To find the maximum, we take the derivative with respect to \( y \):
   \[
   \frac{d}{dy} (2xy - y^2) = 2x - 2y
   \]
   Setting the derivative to zero, we find:
   \[
   2x - 2y = 0 \implies y = x
   \]
   Substituting \( y = x \) back into the expression, we get:
   \[
   2xy - y^2 = 2x^2 - x^2 = x^2
   \]
   Therefore, \( f(x) = x^2 \) satisfies the original equation.

Conclusion:
\[
\boxed{f(x) = x^2 \quad \forall x \in \mathbb{R}}
\]