Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying the functional equation:

$$f(x^2 - y^2) = x f(x) - y f(y)$$

for all real numbers $x$ and $y$.

### Step 1: Explore Simple Cases

Start by setting $x = y$, which gives:

$$f(x^2 - x^2) = x f(x) - x f(x) \implies f(0) = 0$$

### Step 2: Consider $x = 0$ and $y = 0$

1. **Set $x = 0$,** then the equation becomes:

$$f(-y^2) = -y f(y)$$

2. **Set $y = 0$,** then the equation becomes:

$$f(x^2) = x f(x)$$

### Step 3: Analyze $f(x^2) = x f(x)$ and $f(-y^2) = -y f(y)$

From $f(x^2) = x f(x)$, previously derived, we have that for $x = 0$, $f(0) = 0$.

Substituting $y = 0$ in $f(-y^2) = -y f(y)$, reconfirms $f(0) = 0$.

### Step 4: Test Specific Values

Substitute $y = 1$ into the equation $f(x^2 - y^2)$:

$$f(x^2 - 1) = x f(x) - f(1)$$

Setting $f(1) = c$, where $c$ is a constant:

$$f(x^2 - 1) = x f(x) - c$$

### Step 5: Derive General Form

Assume a linear form $f(x) = cx$ for some constant $c$. Substitute into the original equation:

$$f(x^2 - y^2) = c(x^2 - y^2) = x(cx) - y(cy)$$

This simplifies to:

$$cx^2 - cy^2 = cx^2 - cy^2$$

The equation holds true, confirming that $f(x) = cx$ is a solution.

### Conclusion

The only functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the given functional equation are linear functions of the form:

$$\boxed{f(x) = cx}$$

where $c$ is an arbitrary real constant.