Find all functions $f:\mathbb{Q}\rightarrow\mathbb{R}$ such that
    \[
        f(x)^2-f(y)^2=f(x+y)\cdot f(x-y),
    \]
    for all $x,y\in \mathbb{Q}$.

[i]Proposed by Portugal.[/i]
**
   Given the form of \( \mathbb{F}_0 \), we can construct functions \( f \) that satisfy the original equation. One trivial solution is \( f \equiv 0 \). For non-trivial solutions, we can consider functions of the form \( f(x) = k(a^x - a^{-x}) \) for some constants \( k \) and \( a \) in \( \mathbb{R} \) with \( k > 0 \).

The final answer is \( \boxed{ f(x) = k(a^x - a^{-x}) } \) for some constants \( k \) and \( a \) in \( \mathbb{R} \) with \( k > 0 \), or \( f \equiv 0 \).