= Solution
The defining relation $yx=x^4y$ lets us move every occurrence of $y$ to the right of every occurrence of $x$. More generally, repeated <conjugation> gives
$$
yx^r=x^{4r}y,\qquad y^jx^r=x^{4^jr}y^j.
$$
Negative powers cause no problem because $x^{-1}=x^8$ and $y^{-1}=y^2$. Thus every word in the <group presentation> can be put in the required form, reducing powers using $x^9=y^3=e$:
$$
\boxed{g=x^iy^j,\qquad0\le i\le8,\quad0\le j\le2.}
$$
At this stage existence is proved; uniqueness will follow from the concrete <semidirect product> in part (b). Moving $x^r$ past $y^j$ gives the multiplication law
$$
\boxed{(x^iy^j)(x^ry^s)=x^{\,i+4^jr\bmod9}y^{\,j+s\bmod3}.}
$$
The action is consistent with reduction of $j$ modulo $3$ because $4^3\equiv1\pmod9$.
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