Is $\mathbb Q(\zeta_6)=\mathbb {Q}(\zeta_3)$?

Yes, because $\Phi_6$ and $\Phi_3$ are actually $x^2-x+1$ and $x^2+x+1$ respectively. So $\Bbb Q(\zeta_6)$ and $\Bbb Q(\zeta_3)$ have both degree $2$ over $\Bbb Q$ and, since one obviosly contains the other, they are the same extension.

Hint: Note that $\zeta_6=\zeta_3+1$.

Fix an odd integer $n > 2$. If $\alpha$ is a primitive $n$th root of unity then it is easily checked that $-\alpha$ is a primitive $2n$th root of unity.

Also for any algebraic number $\beta$ the number field generated by $\beta$ and $-\beta$ are one and the same.

Now answer for your question can be easily deduced from the above statements which are themselves easy to verify.