Functional analysis through calculus

$\begin{aligned} (x-y)f(x+y) - (x+y) f(x-y) & = 4xy(x^2-y^2) & \small \color{#3D99F6} \text{Let } y = x-1 \\ f(2x-1) - (2x-1)\color{#3D99F6}f(1) & = 4x(x-1)(x-y)(x+y) & \small \color{#3D99F6} \text{Given that }f(1) = -2 \\ f(2x-1) + {\color{#3D99F6}2}(2x-1) & = (4x^2-4x)(2x-1) \\ f(2x-1) & = (2x-1)(4x^2-4x - 2) \\ f{\color{#3D99F6}(2x-1)} & = {\color{#3D99F6}(2x-1)}\left({\color{#3D99F6}(2x-1)}^2 - 3\right) & \small \color{#3D99F6} \text{Replace }2x-1 \text{ with }x. \\ \implies f({\color{#3D99F6}x}) & = {\color{#3D99F6}x}\left({\color{#3D99F6}x}^2 - 3\right) \end{aligned}$

Note that $f(\pm \sqrt 3) = 0$ . The local maximum and minimun within $\left[-\sqrt 3, \sqrt 3\right]$ are when $\dfrac d{dx}f(x) = 3x^2 - 3 = 0$ or $x = \pm 1$ . Since $\dfrac {d^2}{dx}f(x) = 6x$ . $f(-1) = -1\left((-1)^2-3\right) = 2$ is the local maximum and $f(1) = 1\left(1^2-3\right) = -2$ . the minimum. Therefore, $|2-(-2)| = \boxed{4}$ .