<s>[INST] Find all $f$ which satisfies the condition: $f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy$ $$ f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy $$
$$
P(x, 0): f(0)=0 \\
P(0, y): f(f(y)-f(-y))=-y^2\\
P(x, x): f(f(2x))=x^2 \\
P(x, -x): f(-f(2x))=-x^2 \\
P(-x, x): f(-f(-2x))=-x^2 \\
P(-x, -x): f(f(-2x))=x^2 \\
$$
I don't have an idea of the function $f$, but I think we can show that $f$ is an even function.
$$
P(f(x), f(y)): f(f(f(x)+f(y))-\frac{x^2-y^2}{4})=f(x)f(y) \\
P(f(y), f(x)): f(f(f(x)+f(y))+\frac{x^2-y^2}{4})=f(x)f(y) \\
P(f(x), -f(y)): f(\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\
P(f(y), -f(x)): f(-\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\
$$
Please give a perfect process of getting the functions.
 [/INST] Let us rewrite the problem as $f(f(a)-f(b)) = \frac{a^2-b^2}{4}$.
Suppose that there exists a solution;
Claim $1)$ If $f(u) = f(v)$ then $u = \pm v$.
Proof; As you noted (taking $a=b$) we have that $f(0) = 0$. Now
$$0 = f(0) = f(f(u)-f(v)) = \frac{u^2-v^2}{4} \Rightarrow u = \pm v.$$
Claim $2)$ $f(u) = f(-u)$.
Proof; Set $s = f(u)-f(-u)$. We have
$$f(s) = f(f(u)-f(-u)) = 0$$
From $f(0) = 0$ and Claim $1)$ we must have that $s = \pm 0 = 0$.
Now for all $a \in \mathbb{R}$ we have
$$\frac{a^2}{4} = \frac{a^2-0^2}{4} = f(f(a)-f(0)) = f(f(a))$$
$$-\frac{a^2}{4} = \frac{0^2-a^2}{4} = f(f(0)-f(a)) = f(-f(a)) $$
which contradicts claim 2) when we consider $a \neq 0$.
</s>