First of all, we have $f(0) = f(0)^3$ by setting $x = y = z = 0$ in the condition.
Therefore, $f(0) = -1, 0$ or $1$. Now, if $f(0) = 0$, we'll get
$$f(xf(y)) = 0, \forall x, y \in Z$$
by setting $z = 0$. However, this would imply $f(x) = 0, \forall x \in Z$, which is impossible. So it remains to consider the case that $f(0) = -1$ or $f(0) = 1$.

If there is non-zero integer $a$ such that $f(a) = 1$. Then one can set $y = a, z = 0$ to get
$$f(x + a) = f(x), \forall x \in Z$$
which is again impossible because periodic functions defined on integers aren't surjective.

$$f(b^3 - 3b) = f(b)^3 = -1$$

Next, take $x = b^3 - 3b, y = b, z = 0$, we have (Notice $b$ is non-zero)
$$f(4b - b^3) = 1 \Leftrightarrow 4b - b^3 = 0 \Leftrightarrow b = 2, -2$$
Case 1.1: $f(-2) = -1$.
In this case we know
$$f(x) + f(-2 - x) = 0, \forall x \in \mathbb{Z} \qquad (1)$$
Choose $a \in \mathbb{Z}$ satisfying $f(a) = 2$. Set $y = a, z = 0$:
$$f(2x + a) = 2f(x), \forall x \in \mathbb{Z} \qquad (2)$$
In particular, $f(a-4) = 2(-2) = -2$. Use Eq. (1), we get $f(2-a) = 2$. This means
$$f(2x + (2 - a)) = 2f(x) = f(2x + a).$$
Suppose that $2-a \neq a$. Then $f$ could only take finite values on $2\mathbb{Z}+a$. But Eq. (2) then gives that $f$ could only take finite values on integers, which contradicts to the surjectivity of $f$.
Thus, $a$ must be 1 and Eq. (2) becomes
$$f(2x + 1) = 2f(x), \forall x \in \mathbb{Z} \qquad (3)$$
To show $f(x) = x + 1, \forall x \in \mathbb{N}$ ($\mathbb{N}$ is the set of positive integers), we'll apply mathematical induction. Note that
$$f(r - 1) = r \text{ and } f(s - 1) = s \text{ then } f(rs - 1) = rs$$
by setting $x = r, y = s, z = 0$. Consequently, it remains to check that $f(p - 1) = p$ when $p$ is a prime. We have established the based case. Suppose it's true that $f(m - 1) = m$ for all positive integer $m < p$.

Choose $c$ such that $f(c) = p$. If $c \neq -1 \pmod{p}$, take $d \in Z$ so that
$0 \le pd + c < p - 1$. Set $x = d, y = c, z = 0$, we get
$$pd + c + 1 = f(pd + c) = pf(d).$$
This gives $pd + c + 1$ is divisible by $p$, which is absurd! Moreover, if $c + 1$ is a multiple of $2p$, then $c$ is odd. But, it's impossible by Eq. (3).
As the result, $c \equiv p - 1 \pmod{2p}$. Now,
$$f(2px + 2c + 1) = 2f(px + c) = 2pf(x) = pf(2x + 1) = f(2px + p + c)$$
If $2c + 1 \neq p + c$, $f$ can take finite values on $2p\mathbb{Z} + (2p - 1)$. But $2pf(x) = f(2px + p + c)$ implies $f$ can take finite values, a contradiction. We conclude that $2c + 1 = p + c$. In other words, $c = p - 1$, as desired.
Hence, by mathematical induction, we get $f(x) = x + 1, \forall x \in \mathbb{N}$.
Finally,
$$f(x) = x + 1, \forall x \in \mathbb{Z}$$
according to equation (1).

Still, choose $a$ to be the number such that $f(a) = 2$. This time, we have
$$f(a + 4) = 2f(2) = -2 \rightarrow f(-2 - a) = 2.$$
The same argument gives us $-2 - a = a$, i.e. $a = -1$. Again, the base case is checked. Suppose $f(1 - m) = m$ is true for all positive integer $m < p$ where $p$ is a prime. Choose $c \in \mathbb{Z}$ satisfying $f(c) = p$.

Then if $c \neq 1 \pmod{p}$, take $d \in Z$ such that $1 - p < pd + c \leq 0$. Set $x = d, y = c, z = 0$:
$$1 - (pd + c) = f(pd + c) = pf(d).$$
This is also impossible. So $c \equiv 1 \pmod{p}$. Since $c$ is even, it follows that $c \equiv p + 1 \pmod{2p}$.
$$f(2px + 2c - 1) = 2f(px + c) = 2pf(x) = pf(2x - 1) = f(2px - p + c).$$
If $c \neq 1 - p$, then $f$ takes finite values on $2pZ + 1$ and also on $Z$, a contradiction. Therefore, $c = 1 - p$. By mathematical induction,
$$f(1 - x) = x, \forall x \in N.$$
Use $f(x) + f(2 - x) = 0$, we can say
$$f(1 - x) = x, \forall x \in Z \rightarrow f(x) = 1 - x, \forall x \in Z.$$

Choose $e \in Z$ such that $f(e) = 1$ and set $y = e, z = 0$ to get
$$f(x - e) = -f(x) = 0 \rightarrow f(x - 2e) = -f(x - e) = f(x).$$
Because $e$ is non-zero, $f$ is periodic, which is impossible.
It's easy to verify that $f(x) = 1 + x, \forall x \in Z$ and $f(x) = 1 - x, \forall x \in Z$ satisfy the original conditions. In conclusion, they are our answers.