Divisibility in the ring of integers. For example, let $R=\Bbb Z [\sqrt{-5}]$, and I want to explain $3$ does not divide $2-\sqrt{-5}$. I think the following proof will be right:

Suppose $3(a+b\sqrt{-5})=2-\sqrt{-5}$, then taking norms
  $9=N(3)N(a+b\sqrt{-5})=9(a^2+5b^2)$, so that $b=0, a= \pm1$. But $ \pm
> 3 \neq 2-\sqrt{-5}$, contradiction.

But can I explain it just like this? - Since $$\frac{2-\sqrt{-5}}{3} \notin \Bbb Z [\sqrt{-5}]$$ $3$ does not divide $2-\sqrt{-5}$. It seems plausible but any parts of my textbook does not explain like this way.
In other words, is it true that for $x,y \in \Bbb Z [\sqrt{-D}]$, $x$ divides $y$ if and only if $y/x \in \Bbb Z [\sqrt{-D}]$?
 A: Your idea is invalid as it stands, but can be made essentially correct with some more work.

Since $$\frac{2-\sqrt{-5}}{3} \notin \Bbb Z [\sqrt{-5}]$$ $3$ does not divide $2-\sqrt{-5}$.

You're hiding a lot of subtlety in your statement that "$\frac{2-\sqrt{-5}}{3} \notin \Bbb Z [\sqrt{-5}]$".  It is not at all clear what you mean by this.
In particular, what do you mean when you write "$\frac{2-\sqrt{-5}}{3}$"?  Division is not well-defined in the ring $\mathbb{Z}[\sqrt{-5}]$, because not every element has an inverse.  So when you write those symbols $\frac{2-\sqrt{-5}}{3}$, are you referring to an element of $\mathbb{Q}[\sqrt{-5}]$?  Or are you referring to the complex number $\frac{2-i\sqrt{5}}{3}$?
Moreover, how do you conclude from the fact that this complex number is not in the ring $\mathbb{Z}[\sqrt{-5}]$, that $2 - \sqrt{-5}$ is not divisible by $3$?
How is that different from just naming a random number like $\pi$ and saying it is not in $\mathbb{Z}[\sqrt{-5}]$?
How to make your argument rigorous
Your argument can only work with following key observation: the ring $\boldsymbol{\mathbb{Z}[\sqrt{-5}]}$ is contained in the ring of complex numbers, under the mapping (read: injective homomorphism) $a + b\sqrt{-5} \to a + b\sqrt{5} \; i$.
From this observation, we know if $2 - \sqrt{-5}$ were equal to $3 \cdot x$ for some $x \in \mathbb{Z}$, then we would have $2 - \sqrt{5} i = 3 x'$ for the corresponding complex number $x'$.  Then of course that complex number $x'$ would be equal to $\frac{2}{3} - \frac{\sqrt{5}}{3} i$ (since the complex numbers are a field).  But the mapping $a + b\sqrt{-5} \to a + b\sqrt{5} \; i$ does not map anything to $\frac{2}{3} - \frac{\sqrt{5}}{3} i$; therefore, no such $x$ exists.
