= Solution
Take the right <coaction> of the <coordinate Hopf algebra of quantum SL2> on the <quantum plane> to be
$$
\delta(x)=x\otimes a+y\otimes c,\qquad
\delta(y)=x\otimes b+y\otimes d.
$$
It is extended multiplicatively and sends $1$ to $1\otimes1$. The matrix <comultiplication> and <counit> verify the <coaction> identities on $x,y$. To check that it descends to the <quantum plane>, expand
$$
\delta(y)\delta(x)-q\delta(x)\delta(y)
=x^2\otimes(ba-qab)+y^2\otimes(dc-qcd)
+xy\otimes[bc+q da-q ad-q^2cb]=0.
$$
The last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.
For the requested calculation, $ca=qac$ and $yx=qxy$ give
$$
\delta(x)^2=x^2\otimes a^2+(1+q^2)xy\otimes ac+y^2\otimes c^2.
$$
Multiply by $\delta(y)$, put all <monomials> into the order $x^ry^s$, and use $bc=cb$. The answer is
$$
\boxed{\begin{aligned}
\delta(x^2y)={}&x^3\otimes a^2b
+x^2y\otimes\bigl(a^2d+(1+q^2)abc\bigr)\\
&+xy^2\otimes\bigl((1+q^2)acd+q^2bc^2\bigr)
+y^3\otimes c^2d.
\end{aligned}}
$$
A left-<coaction> convention transposes where the matrix coefficients occur; the displayed answer fixes the right <coaction> throughout.
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