Find $f '(0)$ for the function \begin{align} f(x)=\begin{cases} \frac{g(x)}{x^2}, & \text{if }x \not = 0\\ 0, & \text{if }x=0 \end{cases} \end{align}
With \begin{align} g(0)=g'(0) = g''(0) = 0 \\ g'''(0) = 14 \end{align}
Hint: Think of $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$$
Hint: Use the definition and L'Hôpital's rule.
The answer is $\dfrac pq$ with $p,p-q\in\{7,1,8,6,4\}$.