Factorize $x^3-3$ over $\mathbb{Z}_7[y]/\langle y^3-3 \rangle$ 
Factorize $x^3-3$ over $\frac{\mathbb{Z}_7[y]}{\langle y^3-3 \rangle}$

It is clear that $y+\langle y^3-3 \rangle$ is a root of $x^3-3$. Then, I have
$$x^3-3 = (x-y)(x^2+yx+y^2)$$
Can we factorize $x^2+yx+y^2$ further into linear factors?
 A: That quadratic's discriminant is ( all along we do arithmetic modulo $\;7\;$):
$$\Delta=y^2-4y^2=-3y^2=4y^2=(2y)^2$$
and thus its roots are
$$x_{1,2}=\frac{-y\pm2y}2=\begin{cases}\frac{-3y}2=4\cdot(-3y)=2y\\{}\\\frac y2=4y\end{cases}$$
A: Yes, because finite extensions of finite fields are cyclic.
In particular, one can follow SpamIAm's comment and write $x^3-3=(x-y)(x-y^7)\left(x-y^{49}\right)$.
A: Here's one last way to see that your polynomial splits completely.  Just like over $\mathbb{C}$, the polynomial $x^3 - 3$ factors as $(x - \alpha)(x - \zeta \alpha)(x - \zeta^2 \alpha)$ in some algebraic closure, where $\zeta$ is a primitive $3^\text{rd}$ root of $1$ and $\alpha$ is any root.  (Using your notation, $\alpha = y+\langle y^3-3 \rangle$).
However, in this case the base field $\mathbb{F}_7$ already contains $\zeta$: since the group of units $\mathbb{F}_7^\times$ is cyclic and $3 \mid 6 = \#\mathbb{F}_7^\times$, then $\mathbb{F}_7^\times$ contains an element of order $3$, i.e., a primitive $3^\text{rd}$ root of $1$.  Indeed, since $2^3 = 8 = 1$, we see that $2$ is such a root.  Thus the polynomial factors as
$$
x^3 - 3 = (x - \alpha)(x - 2\alpha)(x - 4 \alpha) \, .
$$
