Find all functions $f: R \to R$, such that equality $f (xf (y) - yf (x)) = f (xy) - xy$ holds for all $x, y \in R$.
1. **Initial Substitution and Simplification:**
   Let \( P(x, y) \) denote the assertion of the functional equation:
   \[
   P(x, y): f(xf(y) - yf(x)) = f(xy) - xy
   \]
   Substituting \( y = 0 \) into the equation, we get:
   \[
   P(x, 0): f(xf(0)) = f(0)
   \]
   If \( f(0) \neq 0 \), then for any \( x \), \( xf(0) \) is a constant, implying \( f \) is constant, which contradicts the functional equation. Therefore, \( f(0) = 0 \).

2. **Substitution with \( x = y \):**
   Substituting \( x = y \) into the equation, we get:
   \[
   P(x, x): f(xf(x) - xf(x)) = f(x^2) - x^2 \implies f(0) = f(x^2) - x^2
   \]
   Since \( f(0) = 0 \), it follows that:
   \[
   f(x^2) = x^2 \quad \forall x \in \mathbb{R}
   \]
   This implies \( f(x) = x \) for all \( x \geq 0 \).

3. **Exploring \( f(x) \) for \( x < 0 \):**
   Consider \( P(x, 1) \):
   \[
   P(x, 1): f(xf(1) - f(x)) = f(x) - x
   \]
   Let \( c = f(1) \). Then:
   \[
   f(xc - f(x)) = f(x) - x
   \]
   Similarly, consider \( P(1, x) \):
   \[
   P(1, x): f(f(x) - x) = f(x) - x
   \]
   This implies:
   \[
   f(xc - f(x)) = f(f(x) - x) = f(x) - x
   \]

4. **Behavior for Negative \( x \):**
   Let \( x < 0 \). If \( x \geq f(x) \), then \( x - f(x) \geq 0 \), so:
   \[
   f(x - f(x)) = x - f(x)
   \]
   This implies \( x = f(x) \). Therefore, for \( x < 0 \), \( x \leq f(x) \).

5. **Considering Mixed Signs:**
   Let \( x_1 > 0 \) and \( y_1 < 0 \). Then:
   \[
   P(x_1, y_1): f(x_1 f(y_1) - y_1 f(x_1)) = f(x_1 y_1) - x_1 y_1
   \]
   Since \( f(y_1) \geq y_1 \), we have:
   \[
   f(x_1 f(y_1) - x_1 y_1) = x_1 f(y_1) - x_1 y_1
   \]
   This implies:
   \[
   x f(y) = f(xy) \quad \text{for } x > 0 \text{ and } y < 0
   \]

6. **Exploring \( f(-1) \):**
   Set \( x_1 \rightarrow -y_1 \) and \( y_1 \rightarrow -x_1 \):
   \[
   -y_1 f(-x_1) = f(x_1 y_1) = x_1 f(y_1)
   \]
   For \( x_1 = 1 \):
   \[
   -y_1 f(-1) = f(y_1)
   \]
   Thus, \( f(x) = -x f(-1) \) for \( x < 0 \).

7. **Determining \( f(-1) \):**
   Let \( x_2 < 0 \) and \( y_2 > 0 \):
   \[
   P(x_2, y_2): f(x_2 y_2 - (-x_2) f(-1) y_2) = f(x_2 y_2) - x_2 y_2
   \]
   Since \( f(x_2) \geq x_2 \):
   \[
   f(-1)(f(x_2) y_2 - x_2 y_2) = -x_2 y_2 f(-1) - x_2 y_2
   \]
   This implies \( f(-1) = \pm 1 \).

8. **Final Cases:**
   - If \( f(-1) = -1 \), then \( f(x) = x \) for \( x < 0 \). Thus, \( f(x) = x \) for all \( x \in \mathbb{R} \).
   - If \( f(-1) = 1 \), then \( f(x) = -x \) for \( x < 0 \). Thus, \( f(x) = |x| \) for all \( x \in \mathbb{R} \).

The final answer is \( \boxed{f(x) = x \text{ or } f(x) = |x| \text{ for all } x \in \mathbb{R}} \)