Consider the functional equation that must be satisfied for all functions $f: \mathbb{R} \to \mathbb{R}$:

$$f\left( f(x) - y \right) \cdot f\left( x + f(y) \right) = x^2 - y^2$$

### Step 1: Analyze the structure

1. Initial Observation: The equation is symmetric in a way that resembles the difference of squares:
$$x^2 - y^2 = (x - y)(x + y)$$
The left-hand side involves compositions and products of $f$, indicating the need for careful manipulation or fixed points.

### Step 2: Test simple functions

2. **Assume $f(x) = x$**:
- Substituting $f(x) = x$ into the equation gives:
$$f(x - y) \cdot f(x + y) = (x - y)(x + y) = x^2 - y^2$$
- This equation holds as $f(x-y) = x-y$ and $f(x+y) = x+y$.

### Step 3: Uniqueness Verification

3. Verify uniqueness:
- Assume there exists some $f(x)$ different from $x$. Consider specific $x$ and $y$ to verify potential solutions.

4. Specific Substitutions:
- Use $x = 0$ and $y = 0$ to get:
$$f\left( f(0) \right) \cdot f\left( 0 \right) = 0$$
Suggesting $f(0) = 0$ or imposing conditions on $f(f(0))$.

5. Further Simplification:
- If $f$ is odd, such substitutions confirm:
$$f(f(x) - x) \cdot f(x + f(0)) = x^2$$

6. Deduction from Substitution:
- Assume $x = y$ implies:
$$f(0) \cdot f(x + f(x)) = 0$$
If $f$ is nonzero everywhere, forces $f(0) = 0$.
- Assume $x = -y$:
$$f\left( f(x) + x \right) \cdot f\left( x \right) = 0$$
Recursion imposes that consistent values throughout must align, asserting linear function.

### Conclusion

All trials consistently lead back to the identity function, $f(x) = x$, due to the symmetric structure of the equation. Given any deviation, contradictions arise through substitution symmetry. Therefore, the only solution is:

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