How to show $f^{-1}(D)/f^{-1}(C)\simeq D/C$ by the language of Abelian category? Let $\mathscr C$ be an Abelian category and $f:A\to B$ an epimorphism in $\mathscr C$. 
Let $g:C\to D$ and $h:D\to B$ be monomorphisms in $\mathscr C$, thus $C,D$ are subobjects of $B$. 
Let $f^{-1}(C)$ denote the inverse image of $C$ under $f$ and $f^{-1}(D)$ the inverse image of $D$ under $f$.
How to show $f^{-1}(D)/f^{-1}(C)\simeq D/C$ by the language of Abelian category?
 A: A lemma I find very useful is:

Lemma: The square
  $$ \require{AMScd}\begin{CD}
W @>u >> X
\\ @VVV @VVV
\\ Y @>v >> Z
\end{CD}
$$
  is a pullback square if and only if both of the following conditions are true:
  
  
*
  
*The induced map $\ker(u) \to \ker(v)$ is an isomorphism
  
*The induced map $\operatorname{coker}(u) \to \mathrm{coker}(v)$ is monic
  

For intuition, note the second condition is equivalent to $\mathrm{coker}(u)$ being the image of the composite $X \to Z \to \mathrm{coker}(v)$.
The data you've given can be organized into the following diagram:
$$ \require{AMScd}\begin{CD}
f^{-1}(C) @>>> f^{-1}(D) @>>> A
\\ @VVV @VVV @VV f V
\\ C @>>> D @>>> B
\end{CD} $$
The square on the right and the outer square are pullback squares (by definition of $f^{-1}$); by the pasting lemma this means the square on the left is a pullback square as well.
By the above lemma, $f^{-1}(D) / f^{-1}(C) \to D/C$ is monic.
In an abelian category, pullbacks of epimorphisms are epimorphisms. This implies $f^{-1}(D) \to D$ is epic.
$D \to D/C$ is epic as well. Composites of epics are epic, and so the composite $f^{-1}(D) \to D \to D/C$ is epic as well.
But this composite is equal to $f^{-1}(D) \to f^{-1}(D)/f^{-1}(C) \to D/C$. In any category, if the composite $X \to Y \to Z$ is epic, then so is $Y \to Z$.
We've shown  $f^{-1}(D)/f^{-1}(C) \to D/C$ is both monic and epic. In an abelian category, such morphisms must be isomorphisms.
Note that we did not need to use the fact $g$ and $h$ are monic. (although if we generalize, notation like $D/C$ should be understood as meaning the cokernel of the map $C \to D$)
A: By universal property of cokernels, there exists a morphism $h$ making the following diagram with exact rows commutative:
$$\require{AMScd}$$\begin{CD}
O @>>> f^{-1}(C) @>>> f^{-1}(D) @>>> f^{-1}(D)/f^{-1}(C) @>>> O\\ 
{} @VVV @VVV @VV h V \\ 
O @>>> C @>>> D @>>> D/C @>>>O
\end{CD}
Since $f$ is epic and since epimorphisms are pullback-stable, then $f^{-1}(D)\to D$ is an epimorphism.
Consequently, $h$ is an epimorphisms as well.
Since the left handed-square is a pullback, a diagram chasing show that $h$ is monomorphism, hence an isomorphism.
