1. We need the domain of $R^{-1}$.

   If $R = \{(x,y): x,y\in \mathbb{Z},\ x^2+3y^2\le 8\},$ then $R^{-1} = \{(y,x):(x,y)\in R\}.$

   So, the domain of $R^{-1}$ is the set of all first components in $R^{-1}$, i.e. all values of $y$ for which there exists some integer $x$ such that $x^2+3y^2\le 8.$

2. Thus we must find all integers $y$ satisfying $x^2+3y^2\le 8$ for at least one integer $x$.

   Since $x^2\ge 0$, a necessary condition is $3y^2\le 8.$

   Therefore, $y^2\le \frac{8}{3}.$

   Since $y\in \mathbb{Z}$, this gives $y= -1,\ 0,\ 1.$

3. Check each value:

   • For $y=0$: $x^2+3(0)^2=x^2\le 8,$ possible for example $x=0$.

   • For $y=1$: $x^2+3\le 8 \Rightarrow x^2\le 5,$ possible for example $x=0$.

   • For $y=-1$: $x^2+3\le 8 \Rightarrow x^2\le 5,$ possible for example $x=0$.

   So all three values occur.

4. Hence the domain of $R^{-1}$ is $\{-1,0,1\}.$

5. Comparing with the options:

   • A: $\{0,1\}$ — incorrect
   • B: $\{-2,-1,1,2\}$ — incorrect
   • C: $\{-1,0,1\}$ — correct
   • D: $\{-2,-1,0,1,2\}$ — incorrect

Therefore, the correct answer is Option C.