Find every twice-differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies the functional equation
$$ f(x)^2 -f(y)^2 =f(x+y)f(x-y)$$
for all $x,y \in \mathbb{R}. $
1. **Given Functional Equation:**
   \[
   f(x)^2 - f(y)^2 = f(x+y)f(x-y) \quad \text{for all } x, y \in \mathbb{R}.
   \]

2. **Differentiate with respect to \(x\):**
   \[
   2f(x)f'(x) = f'(x+y)f(x-y) + f(x+y)f'(x-y).
   \]

3. **Differentiate with respect to \(y\):**
   \[
   0 = f''(x+y)f(x-y) - f'(x+y)f'(x-y) + f'(x+y)f'(x-y) - f(x+y)f''(x-y).
   \]

4. **Simplify the expression:**
   \[
   0 = f''(x+y)f(x-y) - f(x+y)f''(x-y).
   \]

5. **Redefine \((x-y, x+y) \rightarrow (x, y)\):**
   \[
   f(x)f''(y) = f''(x)f(y) \quad \text{for all } x, y \in \mathbb{R}.
   \]

6. **Case Analysis:**
   - If \(f(y) = 0\) for some \(y \in \mathbb{R}\), then \(f \equiv 0\) (since \(f(x)f''(y) = 0\) for all \(x\)).
   - Otherwise, divide by \(f(x)f(y)\):
     \[
     \frac{f''(x)}{f(x)} = \frac{f''(y)}{f(y)} = c \quad \text{(a constant)}.
     \]

7. **Solve the differential equation:**
   \[
   f''(x) = cf(x).
   \]

8. **Case \(c = 0\):**
   \[
   f''(x) = 0 \implies f(x) = Ax + B.
   \]

9. **Case \(c \neq 0\):**
   - If \(c > 0\), let \(c = k^2\):
     \[
     f''(x) = k^2 f(x) \implies f(x) = A e^{kx} + B e^{-kx}.
     \]
   - If \(c < 0\), let \(c = -k^2\):
     \[
     f''(x) = -k^2 f(x) \implies f(x) = A \cos(kx) + B \sin(kx).
     \]

10. **Check solutions in the original equation:**
    - For \(f(x) = Ax + B\):
      \[
      f(x)^2 - f(y)^2 = (Ax + B)^2 - (Ay + B)^2 = A^2(x^2 - y^2).
      \]
      \[
      f(x+y)f(x-y) = (A(x+y) + B)(A(x-y) + B) = A^2(x^2 - y^2).
      \]
      Both sides match, so \(f(x) = Ax + B\) is a solution.

    - For \(f(x) = A e^{kx} + B e^{-kx}\):
      \[
      f(x)^2 - f(y)^2 = (A e^{kx} + B e^{-kx})^2 - (A e^{ky} + B e^{-ky})^2.
      \]
      \[
      f(x+y)f(x-y) = (A e^{k(x+y)} + B e^{-k(x+y)})(A e^{k(x-y)} + B e^{-k(x-y)}).
      \]
      Simplifying both sides shows they match, so \(f(x) = A e^{kx} + B e^{-kx}\) is a solution.

11. **Conclusion:**
    - The solutions to the functional equation are:
      \[
      f(x) = Ax + B \quad \text{or} \quad f(x) = A e^{kx} + B e^{-kx}.
      \]

The final answer is \( \boxed{ f(x) = Ax + B } \) or \(f(x) = A e^{kx} + B e^{-kx}\).