what $\frac{x}{y} mod (n)$? assume that $\frac xy \in \mathbb N$ , is it correct that $\frac xy mod (n)=\frac{x \ mod \ n}{y\ mod \ n} mod \ n $ ? if not then how to compute it ?
explination : 
I am dealing with large numbers and I want to compute $\frac xy mod \ n$ , assume that x is very big integer ( can reach $100^{100}$) while $y$ is very small 
 A: I'm assuming that by "$a \mod b$" you mean "the remainder when $a$ is divided by $b$" rather than the (more mathematical) equivalence class definition.  To avoid confusion, I'll instead write "% n" for the remainder modulo $n$, and $\mod n$ for mathematical equivalence modulo $n$.
Since you're assuming $\frac{x}{y} \in \mathbb{N}$, as others have suggested, you can let $x = ky$ for some $k \in \mathbb{N}$.  Then you are asking if 
$$
k \% n = \frac{ky \% n}{y \% n} \% n
$$
This question is actually rather ill-defined, because $ky \% n$ may not be divisible by $y \% n$ at all.  Moreover, if $y \% n = 0$ you will be dividing by zero, which is not good at all.
So, is there another way to compute $\frac{x}{y} \% n$ using the result of $x \% n$?  Well, let's say that $\frac{x}{y} \% n = a$.  Then we should have that $a y \equiv x \mod n$, hence $ay \equiv (x \% n) \mod n$.  From here, you need to compute the inverse of $y$ modulo $x \% n$.  you can then find $a$ (your answer) by
$$
a = \left[ (x \% n) \cdot (\text{inverse of } y \mod x \% n) \right] \% n
$$
Hopefully $n$ is not too large so you can compute the inverse of $y$ relatively easily.  (You specified that $x$ was large and $y$ was small, but didn't say anything about $n$.)
Note that $x \% n$ and $y$ must be relatively prime for this to work; if they aren't, you will simply have to compute $\frac{x}{y}$ directly.  Actually, I'm confused as to why you aren't just computing $\frac{x}{y}$ directly in the first place, and then taking that mod $n$, but I assume there is some reason.
A: $$ x = ky \quad , \frac{x}{y} \pmod n = \frac{x}{y} - n\left \lfloor \frac{x}{ny} \right \rfloor $$ 
$$ \frac{x \pmod n}{y \pmod n} = \frac{x - n\left \lfloor \frac{x}{n} \right \rfloor}{y- n\left \lfloor \frac{y}{n} \right \rfloor } $$
