= Solution
A <monomial> $x^ay^b$ is invariant precisely when $a-b$ is divisible by $n$: pulling back along the generator multiplies it by $\zeta^{a-b}$. Since distinct <monomials> are linearly independent, an invariant polynomial is a linear combination of exactly these invariant monomials. If $a\ge b$, write $a-b=kn$ and obtain
$$
x^ay^b=(x^n)^k(xy)^b.
$$
If $b\ge a$, the analogous expression uses $y^n$ and $xy$. Thus the <polynomial invariant ring> is generated by
$$
u=x^n,\qquad v=y^n,\qquad w=xy,
$$
which satisfy $uv=w^n$.
To check that this is the only relation, reduce each monomial in $\mathbb C[U,V,W]$ using $UV=W^n$ until it has one of the forms
$$
W^c,\qquad U^kW^c,\qquad V^kW^c\quad(k\ge1,\ c\ge0).
$$
Their images are, respectively, $x^cy^c$, $x^{nk+c}y^c$ and $x^cy^{nk+c}$. These are pairwise distinct <monomials> in $x,y$, so no nontrivial linear combination of the reduced forms maps to zero. Therefore
$$
\boxed{\mathbb C[x,y]^{C_n}\cong\mathbb C[U,V,W]/(UV-W^n).}
$$
This identifies the <determinant-one cyclic quotient of the affine plane> with the affine hypersurface $UV=W^n$, and the quotient map is $(x,y)\mapsto(x^n,y^n,xy)$.
Apply the <Jacobian criterion> to $f=UV-W^n$. Its gradient is
$$
\nabla f=(V,U,-nW^{n-1}).
$$
Over $\mathbb C$, with the printed $n\ge3$, this vanishes only at $U=V=W=0$. Every other point is <smooth>. This point is exactly the image of the origin, and its inverse image is only the origin, since $x^n=y^n=0$ forces $x=y=0$. In fact the origin itself is singular: its tangent space has dimension three while the hypersurface has dimension two. Hence \b[the quotient is smooth away from the image of the origin], with that point its unique singularity. Primality is not needed for this calculation; the same argument works for every integer $n\ge2$.
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