In $\mathbb{Z}[\zeta_3]$ is the product of conjugates equal to the conjugate of the product? For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha \cdot \beta)'=\alpha'\cdot \beta'$?
Also,
For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha + \beta)'=\alpha'+ \beta'$?
 A: If $F$ is a field of characteristic different from $2$, and $K$ is an extension with $[K:F]=2$, then the conjugate of $\alpha\in K$ is the image of $\alpha$ under the unique nontidentity element $\sigma\in\mathrm{Gal}(K/F)$ (this agrees with the usual "complex conjugation", since complex conjugation is the unique nonidentity element of $\mathrm{Gal}(\mathbb{C}/\mathbb{F})$). Since $\sigma$ is a field homomorphism, you automatically get both
$$\sigma(\alpha+\beta) = \sigma(\alpha)+\sigma(\beta)\qquad\text{and}\qquad \sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta).$$
In particular, this holds for the unique nonidentity automorphism of $\mathbb{Q}(\zeta_3)$ over $\mathbb{Q}$ (which, as it happens, is complex conjugation). In particular, it holds for elements of $\mathbb{Z}[\zeta_3]$ (which, as it happens, is closed under this automorphism) , so that the restriction of conjugation is in fact also an automorphism of the ring $\mathbb{Z}[\zeta_3]$. 
Of course, you can try defining different kinds of "conjugate". Instead of applying the automorphism of $\mathbb{Q}(\zeta_3)$, you could try defining "conjugate" by $a+b\zeta_3\mapsto a-b\zeta_3$ with $a,b\in\mathbb{Z}$. If you do that, then it's not a ring homomorphism, because
$$(a+b\zeta_3)(x+y\zeta_3) = ax+\zeta_3^2by + (ay+bx)\zeta_3 = ax-by + (ay+bx-by)\zeta_3$$
but 
$$(a-b\zeta_3)(x-y\zeta_3) = ax+\zeta_3^2by - (ay+bx)\zeta_3 = ax-by - (ay+bx+by)\zeta_3$$
so the "conjugate" of the product is not the product of the "conjugates". That's why one uses the field automorphism.
Added. I focused on extensions of degree $2$ to be able to refer to the conjugate. One can of course consider extensions of larger degree or in charactersitic $2$. 
In the general case, the conjugates (plural in general) of $\alpha$ in $K$ are the images of $\alpha$ under automorphisms that lie in $\mathrm{Aut}(K/F)$. Under any particular automorphism $\sigma$ it is still true that $\sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta)$, but if you consider different automorphisms (different notions of "conjugate") of course things go astray. One can also consider the norm, $N_{K/F}$, which maps from $K$ to $F$ by $N_{K/F}(\alpha) = \prod_{\sigma\in G} \sigma(\alpha)$, where $G$ is the Galois group of the Galois closure of $K$ over $F$. The norm is multiplicative: $N_{K/F}(\alpha\beta) = N_{K/F}(\alpha)N_{K/F}(\beta)$; and a very important map. But I doubt this is what you were refering to.
A: If conjugation means mapping $\zeta$ to $\zeta^2=1/\zeta$, then it coincides with complex conjugation, which is a ring homomorphism.
