If $f$ is analytic where $f$ is represented as $f=g.h$ where $g$ is analytic . From here can we conclude that $h$ is analytic? If $f$ is analytic, where  $f$ is represented as $f=g \cdot h,$ where $g$ is analytic. From here can we conclude that $h$ is analytic? 
 A: No take $f\equiv 1$ , $g=x$ and $h=\frac{1}{x}$.
This only works when you allow to observe the limits, without this there are still counter examples.
In fact we can't even conclude that $h$ is continuous anywhere, when $f\equiv 0$ and $g\equiv 0$ you may chose $h$ as any function. To make it not to complicated we just look at the real case (a function is analytic when it locally has a converging taylor series). Chosing $f=0$, $g=0$ and 
$$h=\begin{cases}
0 & x \in \mathbb{R}\setminus\mathbb{Q}\\
1 & x \in \mathbb{Q}\\
\end{cases}$$
gives us a counter example where $h$ is discontinuous everywhere.
As Martin wanted some positive results:
When $f(x_0)\neq 0$ and $g(x_0)\neq 0$ we know that $h$ must be at least locally bounded at $x_0$. 
When we say that $g\neq 0$ then 
$$\frac{f}{g}$$ is complex differentiable which mean that 
$$\frac{f}{g}=h$$ is complex differentiable and hence analytic. 
When $f(x_0)=0$ is a zero of finite order, which means that there exists an $k\in \mathbb{N}$ such that 
$$\lim_{x\to x_0} \frac{f(x)}{(x-x_0)^k}\neq 0$$
And $g$ has a zero of the same order (or lower) in $x_0$ then $h$ is analytic in $x_0$.
Writing $g$ and $f$ as power series gives this result.
