We have $|G|=6$ and by sylow thm we have $n_{3}=1\ n_{2}=1\ or\ 3$ this implies $\exists H \in Syl_{3}(G)$ and $H\lhd G$. Let $K \in Syl_{2}(G)$, then $H\cap K = {1}$ and $HK=G$ and $G\cong K\times_{\tau}H$ for some $\tau:K\rightarrow Aut(H)$
(1)My first question is why $HK=G$?
(2)Why $G\cong K\times_{\tau}H$ for some $\tau:K\rightarrow Aut(H)$?
Let $H=\{1,y,y^2\}$ and $K=\{1,x\}$. And, follow the replies above. $G=\{1,y,y^2,x,xy,xy^2\}$. We know $H\lhd G$ and $K\leq G$ $$\forall g_1,g_2\in G\ g_1=h_1k_1\ g_2=h_2k_2,\ h_1,h_2\in H\ and\ k_1,k_2\in K$$ Observe that: $$g_1g_2=(h_1k_1)(h_2k_2)=h_1k_1h_2(k_1^{-1}k_1)k_2=h_1(k_1h_2k_1^{-1})k_1k_2$$ Let $h_3=xyx^{-1}\in H$, so $h_3=1\ or\ y\ or\ y^2$
Case1: $$If\ h_3=1\Rightarrow y=1 \ contradiction$$
Case2: $$If\ h_3=y\Rightarrow y=y$$ then let $\tau:K\rightarrow Aut(H)\ \tau(1)=id_H\ and\ \tau(x)=id_H\ and\ \tau\ is\ homomorphism.$ $G\cong K\times_\tau H$
Case3: $$If\ h_3=y\Rightarrow h_2=y^2$$ then let $\tau:K\rightarrow Aut(H)\ \tau(1)=id_H\ and\ \tau(x):\{1,y,y^2\}\rightarrow \{1,y^2,y\},\ respectively.\ \tau\ is\ homomorphism.$ $G\cong K\times_\tau H$