Given $f(x)$ is a polynomial of degree $5$.

Also, $f^{\prime }(1)=0;f^{\prime }(-1)=0$.

Also $\because \underset{x\to 0}{\lim }\frac{f(x)}{x^3}=-5$ (fixed and finite)

$\therefore f(0)=0;f^{\prime }(0)=0;f^{\prime \prime }(0)=0$

$\dfrac{f'''(0)}{6}=-5 \;\Rightarrow\; f'''(0)=-30$

Hence, let $f^{\prime }(x)=(ax+b)(x-0)(x-1)(x+1)$

$\Rightarrow\; f'(x)=ax^4+bx^3-ax^2-bx$

$f^{\prime \prime }(x)=4a{x}^3+3b{x}^2-2ax-b$

$\therefore\; f''(0)=0 \Rightarrow b=0$

$f^{\prime \prime \prime }(x)=12a{x}^2+6bx-2a$

$\therefore f^{\prime \prime \prime }(0)=-2a=-30$

$\Rightarrow a=15$

$\therefore\; f'(x)=15x\cdot x\cdot(x-1)(x+1)$

$\Rightarrow\; f'(x)=15x^4-15x^2$

$\Rightarrow\; f(x)=3x^5-5x^3+C$

$\therefore\; f(0)=0 \Rightarrow C=0$

$\therefore\; f(x)=3x^5-5x^3$

$\therefore\; f(2)-f(-2)=112$