Answer
See explanation
Work Step by Step
$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\frac{f(0)}{g(0)}=\frac{0}{g(0)}$$
Since $g$ is the inverse of $f$ it follows:
$$g(f(x))=x$$
$$g(f(0))=0$$
$$g(0)=0$$
so:
$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\frac{f(0)}{g(0)}=\frac{0}{g(0)}=\frac{0}{0}$$ it is an indeterminate form.
Using the l'Hospital's rule it follows:
$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{f'(x)}{g'(x)}=\frac{f'(0)}{g'(0)}$$
we have:
$$f(g(x))=x$$
Differentitate both sides and using the chain rule it follows:
$$g'(x)f'(g(x))=1$$
$$g'(0)f'(g(0))=1$$
$$g'(0)f'(g(0))=1$$
$$g'(0)f'(0)=1$$
$$g'(0)=\frac{1}{f'(0)}$$
so:
$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{f'(x)}{g'(x)}=\frac{f'(0)}{g'(0)}=\frac{f'(0)}{\frac{1}{f'(0)}}=f'(0) \cdot f'(0)=(f'(0))^{2}$$
