To solve the functional equation
$$f(xy)(f(x) - f(y)) = (x-y)f(x)f(y)$$
for all real numbers $x$ and $y$, we proceed as follows:

### Step 1: Analyze the Equation

Given the equation:
$$f(xy)(f(x) - f(y)) = (x-y)f(x)f(y)$$
we can perform a few initial checks. First, consider $x = y$:
$$f(x^2)(f(x) - f(x)) = (x-x)f(x)f(x)$$
This simplifies to $0 = 0$, which is trivially true, but it does not provide any new information.

### Step 2: Substituting Special Values

- **Substituting $y = 0$:**
$$f(0)(f(x) - f(0)) = xf(x)f(0)$$
If $f(0) \neq 0$, then dividing both sides by $f(0)$ gives:
$$f(x) - f(0) = xf(x)$$
Rearranging gives:
$$f(x)(1-x) = f(0)$$
This implies $f(x) = \frac{f(0)}{1-x}$. However, this must hold for all $x$, which is not possible unless $f(x) = 0$ for all $x$.

- **Substituting $x = 0$:**
$$f(0)(f(0) - f(y)) = (-y)f(0)f(y)$$
If $f(0) \neq 0$, then:
$$f(y) = 0 \quad \text{for all } y$$
Hence, $f(x) = 0$ is a possible solution.

### Step 3: Consider Non-zero Solutions

Assume $f(x) \neq 0$ for some non-zero $x$. Substituting $y = 1$, we get:
$$f(x)(f(x) - f(1)) = (x-1)f(x)f(1)$$
Simplifying this gives:
$$f(x)f(x) - f(x)f(1) = (x-1)f(x)f(1)$$
If $f(x) \neq 0$, cancel $f(x)$ from both sides:
$$f(x) - f(1) = (x-1)f(1)$$
Thus,
$$f(x) = f(1)x$$
This implies that $f$ is a linear function of $x$, specifically $f(x) = cx$ for some constant $c$.

### Step 4: Formulate the General Solution

The general solution is:
- If $x \in S$, where $S$ is some subset of $\mathbb{R}$, then $f(x) = f(1)x$.
- If $x \notin S$, then $f(x) = 0$.

Putting it together, we get for some subset $S \subseteq \mathbb{R}$,
$$\boxed{f(x) = f(1)\cdot x \mid x \in S; \quad f(x) = 0 \mid x \notin S}$$

This solution framework satisfies the original functional equation for any choices of sets $S$ by considering both the zero and non-zero functional cases.