<s>[INST] prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$ $f(x)$ and $g(x)$ both differentiable twice at $x = a$ and we know that $f''(a) =g''(a)+f(a)$, $f(a) = g(a) = f'(a) = g'(a) \not = 0$
(we don't know if $f(x)$ and $g(x)$ are differentiable anywhere else except $x=a$)
prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$
I tried to substitute $f^2(x)$ and $g^2(x)$ with a square of taylor expansion of degree 1 (with remainder of degree 2) for example: 
$$f^2(x)= (f(a) + f'(a)x + \frac{f''(\xi)}{2}x^2)^2$$
same for $g^2(x)$.
Now because $f''(x)$ is continuous and since $x \rightarrow a$ we can conclude that $f(\xi) = f(a)$.
Now after simplifying the original expression i get: 
$$\lim\limits_{x \rightarrow a} \frac{f^2(a)(3x^2+2x) + f(a)g''(a) \frac{x^4}{2}}{f^2(a) \cdot (x^2 + \frac{x^2}{2}) + f(a)(x+\frac{x^2}{2})g''(a)x^2 + g''}$$
 [/INST] We can use Taylor series also (instead of L'Hospital's Rule). We have
\begin{align}
f(a + h) &= f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) + o(h^{2})\notag\\
g(a + h) &= g(a) + hg'(a) + \frac{h^{2}}{2}g''(a) + o(h^{2})\notag
\end{align}
We can proceed as follows
\begin{align}
L &= \lim_{x \to a}\frac{f^{2}(x) - g^{2}(x)}{(f(x) - f(a))^{2}}\notag\\
&= \lim_{x \to a}\{f(x) + g(x)\}\cdot\frac{f(x) - g(x)}{(f(x) - f(a))^{2}}\notag\\
&= \{f(a) + g(a)\}\lim_{x \to a}\dfrac{f(x) - g(x)}{\left(\dfrac{f(x) - f(a)}{x - a}\right)^{2}(x - a)^{2}}\notag\\
&= \frac{2f(a)}{\{f'(a)\}^{2}}\lim_{x \to a}\frac{f(x) - g(x)}{(x - a)^{2}}\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{f(a + h) - g(a + h)}{h^{2}}\text{ (putting }x = a + h)\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{1}{h^{2}}\cdot\{f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) - g(a) - hg'(a) - \frac{h^{2}}{2}g''(a) + o(h^{2})\}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f''(a) - g''(a)}{2}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f(a)}{2}\notag\\
&= 1\notag
\end{align}
</s>