A non-Cohen--Macaulay local quotient

Every class in $A$ has a unique expression $f(y)+ax$, where $f\in k\mathrm{⟦}y\mathrm{⟧}$ and $a\in k$: all monomials involving $x$ except $x$ itself belong to $(x^2,xy)$. Thus $x\mathrm{\ne }0$, $x^2=yx=0$, and the projection $A\to k\mathrm{⟦}y\mathrm{⟧}$ has kernel $kx$. An element is a unit precisely when $f(0)\mathrm{\ne }0$: the series $f$ is then invertible by recursive coefficient multiplication, and the square-zero correction is inverted by $1-z$. If $f(0)=0$, its image in the quotient $A\mathrm{/}(x,y)=k$ is zero, so it is not a unit. Hence $A$ is local with maximal ideal $(x,y)$.

Every nonzero series in $k\mathrm{⟦}y\mathrm{⟧}$ is $y^m$ times a unit, by its least nonzero coefficient. Consequently every nonzero ideal in that ring is $(y^m)$, taking the least order among its nonzero elements. For an ideal $J$ of $A$, its projection is therefore principal. If nonzero, lift one generator to $z\in J$; subtracting suitable $k\mathrm{⟦}y\mathrm{⟧}$-multiples of $z$ from elements of $J$ leaves elements of $J\cap kx$. This intersection is either zero or $kx$, since it is a subspace of a one-dimensional $k$-space. Thus $J$ is generated by $z$ and, if needed, $x$; when its projection is zero it is generated by $x$ or is zero. Every ideal is finitely generated, so $A$ is Noetherian by Noetherian commutative rings and modules.

Every prime of $A$ contains the nilpotent $x$, so prime ideals correspond to those of $A\mathrm{/}(x)=k\mathrm{⟦}y\mathrm{⟧}$. The latter has exactly the primes $(0)$ and $(y)$ by the ideal description in step 2.1, and they are distinct. Hence $\dim A=1$.

Every element of $\mathfrak{m}$ annihilates the nonzero $x$. None is therefore an $A$-regular element. Since $A\mathrm{/}\mathfrak{m}=k\mathrm{\ne }0$, the empty sequence is regular and no positive-length regular sequence in $\mathfrak{m}$ exists. Thus $\mathrm{depth}A=0$ by [F1]. The support of the regular module is the whole spectrum, of dimension one, so [F2] shows that $A$ is not Cohen--Macaulay.