Instead of putting so many long comments I thought it would be better to make a small answer. We start with a point $a$ and let us keep notation $(f\circ g)(x) = f(g(x))$. Then we have $$(f\circ g)'(a) = \lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x - a}$$ We are given that $$\lim_{y \to g(a)}\frac{f(y) - f(g(a))}{y - g(a)} = f'(g(a)) = A, \lim_{x \to a}\frac{g(x) - g(a)}{x - a} = g'(a) = B$$ We need to show that $(f\circ g)'(a) = AB$. Clearly we need to distinguish two cases:
1) There is a neighborhood of $a$ in which $g(x) \neq g(a)$ when $x \neq a$. In this case we have $$(f\circ g)'(a) = \lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x - a} = \lim_{x \to a}\frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot\frac{g(x) - g(a)}{x - a} = AB$$
2) Every neighborhood of $a$ contains infinitely many points $x\neq a$ such that $g(x) = g(a)$. Hence it is possible to find a sequence $x_{n} \to a$ such that $x_{n} \neq a$ and $g(x_{n}) = g(a)$. Now limit $B$ exists and hence $$B = \lim_{x \to a}\frac{g(x) - g(a)}{x - a} = \lim_{n \to \infty}\frac{g(x_{n}) - g(a)}{x_{n} - a} = 0$$ Now we need to show that $(f\circ g)'(a) = AB = 0$. Consider the ratio $$F(x, a) = \frac{f(g(x)) - f(g(a))}{x - a}$$ if $g(x) = g(a)$ then $F(x, a) = 0$. If $g(x) \neq g(a)$ then we can write $$F(x, a) = \frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot\frac{g(x) - g(a)}{x - a}$$ and the first factor is near $A$ and second factor is near $B = 0$. So effectively if we have a sufficiently small neighborhood of $a$ then we either have $F(x, a) = 0$ or $F(x, a)$ is very small. Using $\epsilon, \delta$ argument we can show that for any $\epsilon > 0$ there is a $\delta > 0$ such that $|F(x, a)| < \epsilon$ whenever $0 < |x - a| < \delta$. This shows that $\lim_{x \to a}F(x, a) = 0$ i.e. $(f\circ g)'(a) = 0$ as was to shown.
If, for eny $\epsilon > 0 \in \mathbb{R}$, there exists an $n \in \mathbb{Z}$ such that $\forall i > n, |a_i - c| < \epsilon $, then $c$ is defined as the limit of the sequence of $a_i$.
Assuming the limit of sequence $a_i$ is $a > 0$, and the limit of $b_i$ is $b > 0$, we want to prove the limit of $\dfrac{a_i}{b_i} = \dfrac{a}{b}$.
If we apply the limit definitions to $a_i$ and $b_i$ at $\delta < \frac{b}{2}$, how different can $\dfrac{a_i}{b_i}$ be from $\dfrac{a}{b}$?
We apply the limit definition and get a $n_a$ such that $\forall i \ge n_a, |a_i - a| < \delta$, and we get a $n_b$ such that $\forall i \ge n_b, |b_i - b| < \delta$. Let $n' = \max(n_a, n_b)$. We know that for all $i > n', $ both $|a_i - a| < \delta$ and $|b_i - b| < \delta$. We can also say this as $a - \delta \le a_i \le a + \delta$ and $b - \delta \le b_i \le b + \delta$.
So $\dfrac{a - \delta}{b + \delta} \le \dfrac{a_i}{b_i} \le \dfrac{a+\delta}{b-\delta}$. How far do these minimum and maximum bounds stray from $\dfrac{a}{b}$?
$$\begin{align}
\frac{a}{b} - \frac{a-\delta}{b+\delta} & = \frac{a(b+\delta) - (a-\delta)b}{b(b+\delta)}\\
& = \frac{ab + a\delta - ab + \delta b}{b(b+\delta)} \\
& = \frac{(a+b)\delta}{b(b+\delta)}\\
& < \frac{a+b}{b^2}\delta
\end{align}$$
$$\begin{align}
\frac{a + \delta}{b - \delta} - \frac{a}{b} & = \frac{(a + \delta)b - a(b - \delta)}{b(b - \delta)}\\
& = \frac{ab + \delta b - ab + a\delta}{b(b - \delta)}\\
& = \frac{a + b}{b(b - \delta)}\delta\\
& < \frac{a + b}{b\frac{b}{2}}\delta && \text{remember $\delta < \frac{b}{2}$}\\
\frac{a + \delta}{b - \delta} - \frac{a}{b} & < \frac{a + b}{b^2}2\delta
\end{align}
$$
So now we can prove the limit of $\dfrac{a_i}{b_i}$. When we are given an $\epsilon > 0$, we can use the limits of $a_i$ and $b_i$ for the convenient $\delta = \min(\frac{b}{2}, \frac{b^2}{2(a+b)}\epsilon)$ to get $n_a$ and $n_b$. Then we determine $n' = \max(n_a,n_b)$. Then (omitting the trivial $\frac{b}{2} < \frac{b^2}{2(a+b)}\epsilon $ case) we know that for all $i > n'$,
$$\begin{align}
|\frac{a_i}{b_i} - \frac{a}{b}| & \le \frac{2(a+b)}{b^2}\delta \\
& \le \frac{2(a+b)}{b^2}\frac{b^2}{2(a+b)}\epsilon \\
& \le \epsilon
\end{align}$$
And so we fulfill the definition of the limit.
Best Answer
You are only given that $$ \lim_{x\to a} g(x) = B \not=0. $$ Without the quotient rule, you don't even know whether $$ \lim_{x\to a} \frac{1}{g(x)} $$ exists (or that it is equal to $1/B$), so that the use of the product rule in your first step is unjustified.
However, if you prove the statement above, then you can use the product rule to prove the general quotient rule $$ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{A}{B}, $$ as you proposed.
EDIT: Just to be fully clear, here are the statements of the product rule and the quotient rule.
Notice that it is nowhere assumed that $$ \lim_{x\to a} \frac{1}{g(x)} $$ exists. This needs to be proven (which basically amounts to proving the quotient rule).