Find all the continuous bounded functions $f: \mathbb R \to \mathbb R$ such that 
\[(f(x))^2 -(f(y))^2  = f(x + y)f(x - y) \text{  for all } x, y \in \mathbb R.\]
1. **Initial Setup and Zero Function:**
   Let \( P(x, y) \) be the assertion \( f(x)^2 - f(y)^2 = f(x+y)f(x-y) \). 
   
   First, we observe that \( f(x) = 0 \) for all \( x \) is a solution. We now look for non-zero functions. Assume there exists \( u \) such that \( f(u) \neq 0 \).

2. **Odd Function and \( f(0) = 0 \):**
   - Consider \( P\left(\frac{u+x}{2}, \frac{u-x}{2}\right) \):
     \[
     f\left(\frac{u+x}{2}\right)^2 - f\left(\frac{u-x}{2}\right)^2 = f(u)f(x)
     \]
   - Now consider \( P\left(\frac{u-x}{2}, \frac{u+x}{2}\right) \):
     \[
     f\left(\frac{u-x}{2}\right)^2 - f\left(\frac{u+x}{2}\right)^2 = f(u)f(-x)
     \]
   - Adding these two equations, we get:
     \[
     f(u)f(x) + f(u)f(-x) = 0 \implies f(x) + f(-x) = 0 \implies f(-x) = -f(x)
     \]
     Thus, \( f \) is an odd function. Also, setting \( x = 0 \) in the original equation, we get \( f(0)^2 - f(y)^2 = f(y)f(-y) = -f(y)^2 \), which implies \( f(0) = 0 \).

3. **Existence of \( v > 0 \) such that \( f(v) = 0 \):**
   - Assume \( f \) is injective. Since \( f \) is continuous and bounded, it must be monotonic. Let \( \lim_{x \to \infty} f(x) = L \) and \( \lim_{x \to -\infty} f(x) = l \).
   - Consider \( P(x+u, x) \):
     \[
     f(x+u)^2 - f(x)^2 = f(2x+u)f(u)
     \]
     Taking \( x \to \infty \), we get \( Lf(u) = 0 \), so \( L = 0 \). Similarly, taking \( x \to -\infty \), we get \( lf(u) = 0 \), so \( l = 0 \). This contradicts the assumption that \( f \) is not the zero function. Hence, \( f \) is not injective.
   - Therefore, there exist \( b \neq c \) such that \( f(b) = f(c) \). Consider \( P(b, c) \):
     \[
     f(b+c)f(b-c) = 0
     \]
     If \( f(b-c) = 0 \), then \( v = |b-c| \). If \( f(b+c) = 0 \) and \( b+c \neq 0 \), then \( v = |b+c| \). If \( f(b+c) = 0 \) and \( b+c = 0 \), then \( c = -b \) and \( f(b) = f(-b) = -f(b) \), so \( f(b) = 0 \). Thus, \( v = \max(|b|, |c|) \).

4. **Existence of \( a > 0 \) such that \( f(2a) = 0 \) and \( f(x) \neq 0 \) for \( x \in (0, 2a) \) and \( f(x+4a) = f(x) \):**
   - Let \( A = \{ x > 0 \mid f(x) = 0 \} \). \( A \) is non-empty. Let \( v \in A \).
   - Consider:
     \[
     P\left(\frac{x+u}{2} + v, v\right) \implies f\left(\frac{x+u}{2} + v\right)^2 - f\left(\frac{x+u}{2}\right)^2 = 0
     \]
     \[
     P\left(\frac{x-u}{2} + v, v\right) \implies f\left(\frac{x-u}{2} + v\right)^2 - f\left(\frac{x-u}{2}\right)^2 = 0
     \]
     \[
     P\left(\frac{x+u}{2}, \frac{x-u}{2}\right) \implies f\left(\frac{x+u}{2}\right)^2 - f\left(\frac{x-u}{2}\right)^2 = f(x)f(u)
     \]
     \[
     P\left(\frac{x+u}{2} + v, \frac{x-u}{2} + v\right) \implies f\left(\frac{x+u}{2} + v\right)^2 - f\left(\frac{x-u}{2} + v\right)^2 = f(x+2v)f(u)
     \]
   - Subtracting and adding these equations, we get \( f(x+2v) = f(x) \). Let \( a = \frac{1}{2} \inf(A) \). Since \( f \) is continuous, \( f(2a) = 0 \). If \( a = 0 \), \( f \) would be periodic with arbitrarily small periods, implying \( f \) is constant, which is a contradiction. Thus, \( a \neq 0 \), and we have \( f(x+4a) = f(x) \) and \( f(x) \neq 0 \) for \( x \in (0, 2a) \).

5. **Form of the Function:**
   - Since \( f(x) \) is known over \( [0, 2a] \), using \( f(-x) = -f(x) \) and \( f(x+4a) = f(x) \), we can determine \( f \) over all \( \mathbb{R} \).
   - Assume \( f(a) = 1 \). Consider \( P(x, 2a) \):
     \[
     f(x+2a)^2 = f(x)^2
     \]
     For \( x \in (0, 2a) \), \( f(x) > 0 \). Thus, \( f(x+2a) = -f(x) \) for all \( x \). Consider \( P(a-x, a) \):
     \[
     f(a-x)^2 - 1 = f(2a-x)f(-x) = -f(x)^2 \implies f(a-x)^2 + f(x)^2 = 1
     \]
     Let \( g(x) = f(x+a) \). We get:
     \[
     g(0) = 1, \quad g(a) = 0, \quad g(x) > 0 \text{ for } x \in (0, a)
     \]
     \[
     g(2x) = 2g(x)^2 - 1, \quad g(x)^2 + g(a-x)^2 = 1
     \]
   - Let \( h(x) = \cos\left(\frac{\pi x}{2a}\right) \). We have \( g(0) = h(0) \) and \( g(a) = h(a) \). By induction and continuity, \( g(x) = h(x) \) for all \( x \in [0, a] \). Thus, \( f(x) = \sin\left(\frac{\pi x}{2a}\right) \) for \( x \in (0, a) \).
   - Using \( f(-x) = -f(x) \) and \( f(x+2a) = -f(x) \), we get \( f(x) = \sin\left(\frac{\pi x}{2a}\right) \) for all \( x \).

Hence, the general solution is \( f(x) = b \sin\left(\frac{\pi x}{2a}\right) \) where \( b \) and \( a \) are arbitrary real numbers.

The final answer is \( \boxed{ f(x) = b \sin\left(\frac{\pi x}{2a}\right) } \) where \( b \) and \( a \) are arbitrary real numbers.