Find all functions ${f : (0, +\infty) \rightarrow R}$ satisfying $f(x) - f(x+ y) = f \left( \frac{x}{y}\right) f(x + y)$ for all $x, y > 0$.

(Austria)
1. Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   P(x, y): f(x) - f(x + y) = f\left(\frac{x}{y}\right) f(x + y)
   \]
   for all \( x, y > 0 \).

2. Assume \( f(u) = -1 \) for some \( u > 0 \). Then, applying \( P(u, 1) \):
   \[
   f(u) - f(u + 1) = f\left(\frac{u}{1}\right) f(u + 1)
   \]
   Substituting \( f(u) = -1 \):
   \[
   -1 - f(u + 1) = (-1) f(u + 1)
   \]
   Simplifying, we get:
   \[
   -1 - f(u + 1) = -f(u + 1)
   \]
   \[
   -1 = 0
   \]
   This is a contradiction. Therefore, \( f(x) \neq -1 \) for all \( x > 0 \).

3. Rewrite the assertion \( P(x, y) \) as:
   \[
   f(x + y) = \frac{f(x)}{f\left(\frac{x}{y}\right) + 1}
   \]
   Let this be the new assertion \( Q(x, y) \).

4. Subtract \( Q(x, y) \) from \( Q(y, x) \):
   \[
   f(x + y) = \frac{f(x)}{f\left(\frac{x}{y}\right) + 1}
   \]
   \[
   f(x + y) = \frac{f(y)}{f\left(\frac{y}{x}\right) + 1}
   \]
   Equating the two expressions for \( f(x + y) \):
   \[
   \frac{f(x)}{f\left(\frac{x}{y}\right) + 1} = \frac{f(y)}{f\left(\frac{y}{x}\right) + 1}
   \]
   Cross-multiplying, we get:
   \[
   f(x) \left( f\left(\frac{y}{x}\right) + 1 \right) = f(y) \left( f\left(\frac{x}{y}\right) + 1 \right)
   \]
   Simplifying, we obtain:
   \[
   f(x) f\left(\frac{y}{x}\right) + f(x) = f(y) f\left(\frac{x}{y}\right) + f(y)
   \]
   \[
   f(x) f\left(\frac{y}{x}\right) - f(y) f\left(\frac{x}{y}\right) = f(y) - f(x)
   \]

5. Consider \( Q(x + y, z) \):
   \[
   f(x + y + z) = \frac{f(x + y)}{f\left(\frac{x}{z} + \frac{y}{z}\right) + 1}
   \]
   Using the previous result:
   \[
   f(x + y) = \frac{f(x)}{f\left(\frac{x}{y}\right) + 1}
   \]
   Substituting:
   \[
   f(x + y + z) = \frac{\frac{f(x)}{f\left(\frac{x}{y}\right) + 1}}{f\left(\frac{x}{z}\right) + f\left(\frac{y}{z}\right) + 1}
   \]
   Simplifying, we get:
   \[
   f(x + y + z) = \frac{f(x)}{f\left(\frac{x}{z}\right) + f\left(\frac{y}{z}\right) + 1}
   \]

6. Swapping \( x \) and \( y \) and subtracting, we get:
   \[
   f(x) f\left(\frac{y}{z}\right) + f(x) \left( f\left(\frac{y}{x}\right) + 1 \right) = f(y) f\left(\frac{x}{z}\right) + f(y) \left( f\left(\frac{x}{y}\right) + 1 \right)
   \]
   Using the previous result, we conclude:
   \[
   f(x) f\left(\frac{y}{z}\right) = f(y) f\left(\frac{x}{z}\right)
   \]

7. Replacing \( (x, y, z) \) by \( (x, xy, x) \), we get:
   \[
   f(x) f(y) = f(xy) f(1)
   \]
   If \( f(1) = 0 \), this implies:
   \[
   f(x) = 0 \quad \forall x > 0
   \]
   which indeed is a solution.

8. If \( f(1) = c \neq 0 \), we get:
   \[
   f(xy) = \frac{1}{c} f(x) f(y)
   \]
   and \( f(x) \neq 0 \) for all \( x > 0 \).

9. \( Q(x, y) \) becomes:
   \[
   f(x + y) = \frac{f(x) f(y)}{c f(x) + f(y)}
   \]
   Swapping \( x \) and \( y \), we get:
   \[
   c = 1
   \]
   and so:
   \[
   f(xy) = f(x) f(y)
   \]
   and:
   \[
   f(x + y) = \frac{f(x) f(y)}{f(x) + f(y)}
   \]

10. This means that \( \frac{1}{f(x)} \) is both additive and multiplicative, and so is the identity (since all-zero is here impossible).

Therefore, we have two solutions:
\[
\boxed{f(x) = 0 \quad \forall x > 0}
\]
and
\[
\boxed{f(x) = \frac{1}{x} \quad \forall x > 0}
\]