Let's determine all twice differentiable real functions $f$ which satisfy the functional equation

(1)

Substituting first $y = 0$ in (1) we see that $f(x)^2 = f(x)^2\!-\!f(0)^2$ or $f(0) = 0$ . The substitution $x = 0$ gives $f(y)f(-y) = -f(y)^2$ , whence $f(-y) = -f(y)$ . So $f$ is an odd function.

We differentiate both sides of (1) with respect to $y$ and the result with respect to $x$ : $$f'(x\!+\!y)f(x\!-\!y)\!-\!f'(x\!-\!y)f(x\!+\!y) = -2f(y)f'(y)$$ $$f''(x\!+\!y)f(x\!-\!y)\!+\!f'(x\!-\!y)f'(x\!+\!y)\!-\! f''(x\!-\!y)f(x\!+\!y)\!-\!f'(x\!+\!y)f'(x\!-\!y) = 0$$ The result is simplified to $f''(x\!+\!y)f(x\!-\!y) = f''(x\!-\!y)f(x\!+\!y)$ , i.e. $$f''(x\!+\!y)/f(x\!+\!y) = f''(x\!-\!y)/f(x\!-\!y).$$ Denoting $x\!+\!y := u$ , $x\!-\!y := v$ we obtain the equation $$\frac{f''(u)}{f(u)} = \frac{f''(v)}{f(v)}$$ for all real values of $u$ and $v$ . This is not possible unless the proportion $\frac{f''(u)}{f(u)}$ has a constant value, independent on $u$ . Thus the second order homogeneous linear differential equation $f''(t)/f(t) = \pm{k}^2$ or $$f''(t) = \pm{k}^2f(t),$$ with $k$ some constant, is valid.

There are three cases:

1. $k = 0$ . Now $f''(t) \equiv 0$ and consequently $f(t) \equiv Ct$ . If one especially chooses the constant $C$ equal to 1, the solution is the identity function $f:t\mapsto{t}$ . This yields from (1) the well-known ``memory formula'' $$(x\!+\!y)(x\!-\!y) = x^2\!-\!y^2.$$
2. $f''(t) = -k^2f(t)$ with $k\neq 0$ . According to the oddness one obtains for the general solution the sine function $f:t\mapsto{C\sin{kt}}$ . The special case $C = k = 1$ means in (1) the formula $$\sin(x\!+\!y)\sin(x\!-\!y) = \sin^2x-\sin^2y,$$ which is easy to verify by using the addition and subtraction formulae of sine.
3. $f''(t) = k^2f(t)$ with $k\neq{0}$ . According to the oddness we obtain for the general solution the hyperbolic sine function $f:t\mapsto{C\sinh{kt}}$ . The special case $C = k = 1$ gives from (1) the formula $$\sinh(x\!+\!y)\sinh(x\!-\!y) = \sinh^2x-\sinh^2y.$$

The solution method of (1) is due to andik and perucho.