= Solution
The class-two qualifier is essential: the presentation is in the variety of <two-step nilpotent groups>, so $\gamma_3(G)=1$ is included. Use the <group commutator> convention $[a,b]=a^{-1}b^{-1}ab$ and put $c=[x,y]$. Centrality and the allowed commutator identities give $[x^3,y]=c^3$, so an ordinary finite presentation of the same <group> is
$$
G=\langle x,y,c\mid c=[x,y],\ [c,x]=[c,y]=1,\ c^3=1\rangle.
$$
The <commutator subgroup> is generated by $c$: modulo it the two generators commute, and centrality makes its normal closure simply its cyclic subgroup. To prove that $c$ has exactly order three and that no hidden relations remain, construct the <group> on triples
$$
E=\mathbb Z\times\mathbb Z\times(\mathbb Z/3\mathbb Z),\qquad
(a,b,r)(a',b',r')=(a+a',b+b',r+r'-a'b).
$$
The bilinear cross term makes multiplication associative: both bracketings contribute $-a'b-a''b-a''b'$ to the last coordinate. The identity is $(0,0,0)$ and the inverse is $(-a,-b,-r-ab)$. The elements $X=(1,0,0)$ and $Y=(0,1,0)$ have central commutator $(0,0,1)$ of order three, and generate $E$.
Conversely $yx=xyc^{-1}$ permits <word collection in a two-step nilpotent group>, expressing every element as $x^ay^bc^r$ with $r$ taken modulo three. The map $x\mapsto X$, $y\mapsto Y$, $c\mapsto(0,0,1)$ sends that collected word to $(a,b,r)$. Equality of such coordinates is therefore necessary as well as sufficient. This proves the unique normal form and the requested finite presentations:
$$
\boxed{G/\gamma_2(G)=\langle\bar x,\bar y\mid[\bar x,\bar y]=1\rangle\cong\mathbb Z^2,\qquad
\gamma_2(G)=\langle c\mid c^3=1\rangle\cong C_3.}
$$
To solve the <word problem for a group>, first total the exponents of $x$ and $y$; they give the quotient coordinates $a,b$. If either is nonzero, the word is nonidentity. For the central coordinate, scan the word maintaining $(a,b,r)$ initially zero. Reading $x^\varepsilon$ changes $(a,r)$ to $(a+\varepsilon,r-\varepsilon b)$, while reading $y^\varepsilon$ changes $b$ to $b+\varepsilon$, with $\varepsilon=\pm1$. Reduce $r$ modulo three. Hence
$$
\boxed{w=1\ \Longleftrightarrow\ a=b=0\ \text{and}\ r\equiv0\pmod3.}
$$
For example, $[x^2,y][x,y]=c^2c=1$, whereas $yx^2y^{-1}x^{-2}=c^{-2}=c\ne1$. Both examples have zero abelian quotient coordinates; the second finite presentation detects their different central coordinates. The displayed one-relator presentation without the class-two laws would be a different group.
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