To show that $$\mathbb{Q}(2^{1/x}, 2^{1/y}) \subseteq \mathbb{Q}(2^{1/{xy}})$$ knowing that $(x,y)=1$, $x, y \in \mathbb{N}$ can we do the following??
$$2^{\frac{1}{x}}=2^{\frac{y}{xy}}=\left ( 2^{\frac{1}{xy}} \right )^y \in \mathbb{Q}(2^{1/{xy}})$$
$$2^{\frac{1}{y}}=2^{\frac{x}{xy}}=\left ( 2^{\frac{1}{xy}} \right )^x \in \mathbb{Q}(2^{1/{xy}})$$
Therefore, $\mathbb{Q}(2^{1/x}, 2^{1/y}) \subseteq \mathbb{Q}(2^{1/{xy}})$.
Is this correct??
It is indeed correct.
Nevertheless, you have not used the condtion $(x,y)=1$. In fact, you do not need it, as $$ \mathbb Q(2^{1/x},2^{1/y})\subset\mathbb Q(2^{1/xy}), $$ even in the case when $x$ and $y$ are not relatively prime.