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: 
$$\lim\limits_{x \rightarrow a} \frac{f^2(a)(3x^2+2x) + f(a)g''(a) \frac{x^4}{2}}{f^2(a) \cdot (x^2 + \frac{x^2}{2}) + f(a)(x+\frac{x^2}{2})g''(a)x^2 + g''}$$

If you use the Taylor polynomial, the answer is easy and short. Note
$$ f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(\xi)(x-a)^2, g(x)=g(a)+g'(a)(x-a)+\frac{1}{2}g''(\eta)(x-a)^2$$
where $\xi$ and $\eta$ are between $a$ and $x$.
So
\begin{eqnarray}
\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} &=&\lim\limits_{x \rightarrow a} (f(x)+g(x))\frac{f(x)- g(x)}{(f(x) -f(a))^2}\\
&=&\lim\limits_{x \rightarrow a} 2f(a)\frac{\frac{1}{2}(f''(\xi)- g''(\eta))(x-a)^2}{[f'(a)(x-a)+O((x-a)^2)]^2}\\
&=&\lim\limits_{x \rightarrow a} f(a)\frac{f''(\xi)- g''(\eta)}{[f'(a)+O((x-a))]^2}\\
&=&f(a)\frac{f(a)}{f^2(a)}\\
&=&1.
\end{eqnarray}
