Every homomorphic image of abelian group is abelian but converse need not be true. Every  homomorphic image  of abelian  group is  abelian but converse  need  not be  true.

My  exercise is  to give an counterexample  that  converse  need not be true.

My attempt  :  I tried to construct a  mapping  $f : S_3 \rightarrow  \frac{S_3}{A_3}$ defined by $f(x) = x$ where $ x \in S_3$ 

I thinks  this  will  be  counter example  but here $S_3$  is non- abelian. 

Is  jt true?
I would have written $f(x)=xA_3$, as that's what the elements of $S_3/A_3$ look like. But otherwise, yeah, that's a good example. This $f$ (and all other homomorphisms constructed from a group to a quotient of that group in this way) is called the canonical quotient homomorphism. Because it's the most obvious way to map a group into a quotient.

For an even simpler example, take any non-abelian group and map it to the trivial group (or to the identity element of your favourite group).
Another  counter example  from trancelocation answer . $$\det:  GL(n,R) \rightarrow \{-1,1\}$$