1. We need the range of

$f(x)=\frac{x^2}{1-x^2}, \qquad x\in \mathbb R\setminus\{1,-1\}.$

Since the function is surjective onto $A$, we must have

$A=\text{Range}(f).$

2. Let

$y=\frac{x^2}{1-x^2}.$

We solve for $x^2$ in terms of $y$:

$y(1-x^2)=x^2$ $y-yx^2=x^2$ $y=x^2(1+y)$ $x^2=\frac{y}{1+y}, \qquad y\neq -1.$

Now, since $x^2\ge 0$, we need

$\frac{y}{1+y}\ge 0.$

3. Solve the inequality

$\frac{y}{1+y}\ge 0, \qquad y\neq -1.$

Critical points are $y=-1$ and $y=0$.

Checking intervals:

• For $y<-1$: numerator $<0$, denominator $<0$, so ratio $>0$.
• For $-1<y<0$: numerator $<0$, denominator $>0$, so ratio $<0$.
• For $y\ge 0$: ratio $\ge 0$.

Hence possible values are

$y\in (-\infty,-1)\cup [0,\infty).$

4. Check endpoints:

• $y=-1$ is impossible since denominator $1+y=0$.
• $y=0$ is possible: $x=0$ gives $f(0)=0$.

So the range is

$(-\infty,-1)\cup [0,\infty).$

This can be written as

$\mathbb R\setminus [-1,0).$

5. Compare with options:

• A: $\mathbb R-(-1,0)=(-\infty,-1]\cup[0,\infty)$ includes $-1$, wrong.
• B: $\mathbb R-\{-1\}$ includes values in $(-1,0)$, wrong.
• C: $\mathbb R-[ -1,0)=(-\infty,-1)\cup[0,\infty)$, correct.
• D: $[0,\infty)$ misses values less than $-1$, wrong.

Therefore,

$A=\mathbb R\setminus[-1,0).$