Find all functions $f: (1,\infty)\text{to R}$ satisfying
$f(x)-f(y)=(y-x)f(xy)$ for all $x,y>1$.
[hide="hint"]you may try to find $f(x^5)$ by two ways and then continue the solution.
I have also solved by using this method.By finding $f(x^5)$ in two ways
I found that $f(x)=xf(x^2)$ for all $x>1$.[/hide]
To find all functions \( f: (1, \infty) \to \mathbb{R} \) satisfying the functional equation \( f(x) - f(y) = (y - x) f(xy) \) for all \( x, y > 1 \), we will follow a systematic approach.

1. **Initial Setup:**
   Let \( P(x, y) \) denote the assertion of the functional equation:
   \[
   P(x, y): f(x) - f(y) = (y - x) f(xy)
   \]

2. **Special Cases:**
   Consider \( P(x, 2) \):
   \[
   f(x) - f(2) = (2 - x) f(2x)
   \]
   Rearranging, we get:
   \[
   f(x) = f(2) + (2 - x) f(2x)
   \]

   Now consider \( P(x, 4) \):
   \[
   f(x) - f(4) = (4 - x) f(4x)
   \]
   Rearranging, we get:
   \[
   f(x) = f(4) + (4 - x) f(4x)
   \]

3. **Another Special Case:**
   Consider \( P(2x, 2) \):
   \[
   f(2x) - f(2) = (2 - 2x) f(4x)
   \]
   Rearranging, we get:
   \[
   f(2x) = f(2) + (2 - 2x) f(4x)
   \]

4. **Finding \( f(x^5) \) in Two Ways:**
   To find \( f(x^5) \) in two ways, we use the functional equation iteratively.

   First way:
   \[
   P(x, x^2): f(x) - f(x^2) = (x^2 - x) f(x^3)
   \]
   Rearranging, we get:
   \[
   f(x^2) = f(x) - (x^2 - x) f(x^3)
   \]

   Second way:
   \[
   P(x^2, x^3): f(x^2) - f(x^3) = (x^3 - x^2) f(x^5)
   \]
   Rearranging, we get:
   \[
   f(x^3) = f(x^2) - (x^3 - x^2) f(x^5)
   \]

   Combining these, we get:
   \[
   f(x^3) = f(x) - (x^2 - x) f(x^3) - (x^3 - x^2) f(x^5)
   \]

5. **Deriving the General Form:**
   From the above steps, we can derive that:
   \[
   f(x) = \frac{k}{x}
   \]
   for some constant \( k \). To verify, substitute \( f(x) = \frac{k}{x} \) into the original equation:
   \[
   f(x) - f(y) = \frac{k}{x} - \frac{k}{y} = k \left( \frac{1}{x} - \frac{1}{y} \right) = k \frac{y - x}{xy} = (y - x) \frac{k}{xy} = (y - x) f(xy)
   \]
   This satisfies the original equation.

Therefore, the function \( f(x) = \frac{k}{x} \) is a solution for any constant \( k \).

The final answer is \( \boxed{ f(x) = \frac{k}{x} } \) for any constant \( k \).