= Solution
For the given <group homomorphism> $\varphi:H\to\operatorname{Aut}(N)$, the external <semidirect product> has underlying <Cartesian product> $N\times H$ and operation
$$
\boxed{(n,h)(m,k)=\bigl(n\varphi(h)(m),hk\bigr).}
$$
The notation $\operatorname{Aut}(N)$ denotes the <automorphism group> of $N$. Closure follows because $\varphi(h)$ is an <automorphism> of $N$. To check <associativity>, calculate both bracketings:
$$
\begin{aligned}
((n,h)(m,k))(r,l)
&=\bigl(n\varphi(h)(m)\varphi(hk)(r),hkl\bigr),\\
(n,h)((m,k)(r,l))
&=\bigl(n\varphi(h)(m\varphi(k)(r)),hkl\bigr)\\
&=\bigl(n\varphi(h)(m)\varphi(h)\varphi(k)(r),hkl\bigr).
\end{aligned}
$$
They agree because $\varphi(hk)=\varphi(h)\varphi(k)$. The identity is $(e_N,e_H)$, using $\varphi(e_H)=\mathrm{id}_N$, and the two-sided inverse is
$$
(n,h)^{-1}=\bigl(\varphi(h^{-1})(n^{-1}),h^{-1}\bigr).
$$
Indeed multiplication in either order gives the identity, using $\varphi(h)\varphi(h^{-1})=\mathrm{id}_N$. This verifies the <group axioms>.
For the specified <group presentation>, take $N=C_9=\langle X\rangle$ and $H=C_3=\langle Y\rangle$. The rule $\alpha(X)=X^4$ defines an <automorphism> because $4$ is coprime to $9$. Furthermore $4^3\equiv1\pmod9$, so $Y\mapsto\alpha$ defines the required <group homomorphism> $C_3\to\operatorname{Aut}(C_9)$. In the resulting <semidirect product> $S$, the canonical generators satisfy
$$
X^9=Y^3=e,\qquad YXY^{-1}=X^4.
$$
Consequently the <group presentation> defines a surjective <group homomorphism> $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying <Cartesian product> gives $|S|=27$, whereas part (a) gives $|G|\le27$. Surjectivity forces equality, so the map is an <isomorphism>:
$$
\boxed{G\cong C_9\rtimes_{\alpha}C_3,\qquad\alpha(X)=X^4,\qquad|G|=27.}
$$
This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has <element order> exactly $9$.
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