= Solution
A <group homomorphism> $D:G_1\to G_2$ satisfies $D(gh)=D(g)D(h)$ for all $g,h\in G_1$. Its <group kernel> is
$$
\ker D=\{g\in G_1:D(g)=e_{G_2}\}.
$$
For the matrices, put $X=R(x)$, $Y=R(y)$ and $z=e^{2\pi i/3}$. Since $z^3=1$, $z\ne1$ and $1+z+z^2=0$, direct <matrix multiplication> gives
$$
X^2=I_2,\qquad Y^2=\begin{pmatrix}z^2&0\\0&z\end{pmatrix},\qquad Y^3=I_2,
$$
$$
XY=\begin{pmatrix}0&z^2\\z&0\end{pmatrix},\qquad
XY^2=\begin{pmatrix}0&z\\z^2&0\end{pmatrix}=YX,\qquad
Y^2X=XY.
$$
These verify the listed matrices and the defining relations, with $R(e)=I_2$. To check every product explicitly through a single calculation, write elements in the unique form $x^\epsilon y^j$, with $\epsilon=0,1$ and $j=0,1,2$. The relations give
$$
(x^\epsilon y^j)(x^\eta y^k)=x^{\epsilon+\eta}y^{(-1)^\eta j+k}.
$$
The displayed matrices obey $Y^jX=XY^{-j}$, so their product has exactly the same form:
$$
(X^\epsilon Y^j)(X^\eta Y^k)=X^{\epsilon+\eta}Y^{(-1)^\eta j+k}=R\big((x^\epsilon y^j)(x^\eta y^k)\big).
$$
All exponents are reduced modulo two or three respectively. Thus this verifies all 36 products, and $R$ is a two-dimensional complex <group representation>. It is given to be an <irreducible representation>.
For a one-dimensional complex <group representation> $U$, let $u=U(x)$ and $v=U(y)$, which are nonzero complex numbers. The relations require
$$
u^2=1,\qquad v^3=1,\qquad vu=uv^2.
$$
Since scalar multiplication commutes and $u\ne0$, the last relation gives $v=v^2$, hence $v=1$. Then $u=\pm1$. The choice $u=1$ is the <trivial representation>; the unique nontrivial choice is the <sign representation>:
$$
\boxed{U(e)=U(y)=U(y^2)=1,\qquad U(x)=U(xy)=U(xy^2)=-1.}
$$
Its <group kernel> is
$$
\boxed{\ker U=\{e,y,y^2\}.}
$$
A <faithful representation> is injective and has trivial kernel. Here the kernel has three elements, so \b[$U$ is not faithful].
Both one-dimensional representations are <irreducible representations>, and they are inequivalent to each other and to the given two-dimensional one. The <sum of squares of irreducible representation dimensions> gives
$$
1^2+1^2+2^2=6=|S_3|.
$$
These exhaust all complex <irreducible representations>. Their number equals the number of <conjugacy classes>, so there are \b[three conjugacy classes]. Explicitly they are
$$
\boxed{\{e\},\qquad\{y,y^2\},\qquad\{x,xy,xy^2\}.}
$$
The last two sets consist of the 3-cycles and transpositions respectively.
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