Given a twice differentiable function $f:(-3,3)\to \mathbb{R}$ such that $$\frac{(x-3)f''(x)-3f'(x)}{x^4}+\frac{12f'(x)}{x^5}\le 2.$$ Find the maximum value of $\int_{-3}^3 \frac{f(x)}{x^3}\;dx$.
I start by constructing $g(x) = \frac{f(x)}{x^3}$ and get the form of $$g''(x) =\frac{(x-3)f''(x)-3f'(x)}{x^4}+\frac{12f(x)}{x^5}.$$ Therefore, $g''(x)\le 2+12\cdot \frac{f(x)-f'(x)}{x^5}$ and I had no progress from here. Any help greatly appreciated.
Edit: It should ask for the maximum value, not the minimum.