The following is from "Calculus 4th Edition" by Michael Spivak.
THEOREM 9(THE CHAIN RULE)
If $g$ is differentiable at $a$, and $f$ is differentiable at $g(a)$, then $f\circ g$ is differentiable at $a$, and $$(f\circ g)^{'}(a)=f^{'}(g(a))\cdot g^{'}(a).$$
My wrong argument is here:
Assume that $g$ be differentiable at $a$.
Assume that $f$ be differentiable at $g(a)$.
We consider the case in which $f^{'}(g(a))\cdot g^{'}(a)\neq 0$.
We assume that for any positive real number $\epsilon$, there exists $h$ such that $-\epsilon<h<\epsilon$ and $g(a+h)-g(a)=0$.
The following equation holds when $g(a+h)-g(a)\neq 0$.
$$\frac{f(g(a+h))-f(g(a))}{h}=\frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)}\cdot\frac{g(a+h)-g(a)}{h}.$$
If $|h|$ is very small, then $|g(a+h)-g(a)|$ is also very small since $g$ is continuous at $a$.
If $|h|$ is very small and $g(a+h)-g(a)\neq 0$, then $\frac{f(g(a+h))-f(g(a))}{h}$ is very close to $f^{'}(g(a))\cdot g^{'}(a)$ since $f$ is differentiable at $g(a)$ and $g$ is differentiable at $a$.
If $|h|$ is very small and $|g(a+h)-g(a)|=0$, then $\frac{f(g(a+h))-f(g(a))}{h}=0$.
$f^{'}(g(a))\cdot g^{'}(a)\neq 0$ by assumption.
So, $\lim_{h\to 0} \frac{f(g(a+h))-f(g(a))}{h}$ cannot exist.
What is wrong with my above argument?