 Solution

The identity is one conjugacy class. In $N\setminus \{1\}$, conjugation by $c$ pairs each element with its inverse, giving four classes represented by $a,b,ab,a{b}^2$, each of size two. All nine elements of $Nc$ are conjugate, since conjugation by $n\in N$ changes the reflection component by $n^2$ and squaring is a bijection of $N$.

There are two linear characters: the trivial character and the character equal to one on $N$ and minus one on $Nc$. Write $\omega =e^{2\pi i/3}$. The remaining four irreducible characters of a representation are induced from the nontrivial characters $a^i{b}^j\mapsto {\omega }^{ri+sj}$ of $N$, pairing $(r,s)$ with $(-r,-s)$. They are exactly the two-dimensional representations already proved irreducible. Taking representatives $(1,0),(0,1),(1,1),(1,2)$ yields the character tableIndeed the induced value on $a^i{b}^j$ is ${\omega }^{ri+sj}+{\omega }^{-(ri+sj)}$, equal to two or minus one, and is zero off $N$. The squared dimensions sum to $1+1+4\cdot 4=18$, so this is the complete table.

An element of $N$ centralizes $c$ only if it equals its inverse, hence only if it is the identity. No reflection centralizes $a$. Thus $Z(E)=\{1\}$ is cyclic, but the two linear representations kill $N$ and each two-dimensional irreducible has a nontrivial kernel of a group homomorphism of order three. Therefore a cyclic centre does not imply a faithful irreducible representation.