Let $H$ be a subgroup of $G$ and $x,y \in G$. Show that $x(Hy)=(xH)y.$
You are entirely correct; there isn't much to be shown, and you've shown it.
You're correct.
This can be done in a few lines:
$$\begin{align} x(Hy)&=\{xh'\mid h'\in Hy\}\\ &=\{x(hy)\mid h\in H\}\\ &=\{(xh)y\mid h\in H\}\\ &=\{h''y\mid h''\in xH\}\\ &=(xH)y. \end{align}$$
If $w\in x(Hy),$ then for some $v\in Hy,$ $w=xv.$ Since $v\in Hy,$ for some $h\in H,$ $v=hy.$ So $w = xhy.$ So for some $u\in xH,$ $w=uy.$ Thus $w\in (xH)y.$
Therefore $x(Hy)\subseteq (xH)y.$
The inclusion in the other direction can be shown in the same way.