If $$R = \{(x, y) : x, y \in Z, x^2 + 3y^2 \le 8\}$$ is a relation on the set of integers $$Z$$, then the domain of $$R^{-1}$$ is:

First, recall that the inverse relation $$R^{-1}$$ is obtained by interchanging the coordinates of every ordered pair of $$R$$. Consequently, every first component (the “domain”) of $$R^{-1}$$ comes from a second component of some ordered pair in $$R$$. In other words,

$$\text{Domain}(R^{-1})=\text{Range}(R).$$

Therefore we must determine all integer values of $$y$$ for which there exists at least one integer $$x$$ satisfying the inequality that defines $$R$$, namely

$$x^2+3y^2\le 8.$$

We systematically test every integer value of $$y$$ whose square cannot make the left-hand side exceed 8.

Let us examine successive values of $$y$$.

1. Take $$y=0$$. We have $$y^2=0$$, so the inequality becomes

$$x^2+3(0)^2\le 8 \;\Longrightarrow\; x^2\le 8.$$

The integer squares not exceeding 8 are $$0,1,4$$. Thus the corresponding integer values of $$x$$ are

$$x=\,-2,\,-1,\,0,\,1,\,2.$$

Since at least one such $$x$$ exists, $$y=0$$ appears in the range of $$R$$.

2. Next take $$y=1$$. We have $$y^2=1$$, giving

$$x^2+3(1)^2\le 8 \;\Longrightarrow\; x^2+3\le 8 \;\Longrightarrow\; x^2\le 5.$$

The integer squares not exceeding 5 are again $$0,1,4$$, so

$$x=\,-2,\,-1,\,0,\,1,\,2.$$

Hence $$y=1$$ also lies in the range of $$R$$.

3. Because the expression depends on $$y^2$$, the same result holds for $$y=-1$$. Indeed, $$(-1)^2=1$$, so the identical calculation shows at least one suitable $$x$$ exists. Thus $$y=-1$$ is present in the range of $$R$$.

4. Now consider $$y=2$$ or $$y=-2$$. Here $$y^2=4$$, leading to

$$x^2+3(4)\le 8 \;\Longrightarrow\; x^2+12\le 8 \;\Longrightarrow\; x^2\le -4.$$

The inequality $$x^2\le -4$$ has no integer solution because a square is always non-negative. Therefore $$y=\pm2$$ do not belong to the range of $$R$$.

5. For any $$|y|\ge 2$$, the value of $$3y^2$$ is at least 12, which already exceeds 8, so no integer $$x$$ can satisfy the defining inequality. Thus no further $$y$$ values are possible.

Collecting all successful $$y$$ values, we obtain

$$\text{Range}(R)=\{-1,\,0,\,1\}.$$

Therefore

$$\text{Domain}(R^{-1})=\{-1,\,0,\,1\}.$$

Hence, the correct answer is Option D.