Step 1: Set the output equal to $y$.
Let $y=\dfrac{x^2}{1-x^2}$. To find the range we ask: for which $y$ can we find a real $x$ with $x\ne\pm1$?
Step 2: Solve for $x^2$.
Cross-multiplying, $y(1-x^2)=x^2$, so $y=x^2+yx^2=x^2(1+y)$. Hence $x^2=\dfrac{y}{1+y}$.
Step 3: Impose that $x^2$ is non-negative.
Since $x^2\ge 0$ for real $x$, we need $\dfrac{y}{1+y}\ge 0$.
Step 4: Solve the sign inequality.
The fraction $\dfrac{y}{1+y}$ is non-negative when both parts have the same sign. It holds for $y\ge 0$, and also for $y<-1$ (there both numerator and denominator are negative). At $y=-1$ the denominator is zero, so that is excluded.
Step 5: Check the boundary $y=0$.
If $y=0$ then $x^2=0$, giving $x=0$, which is allowed. So $y=0$ is in the range.
Step 6: Collect the range.
Combining, the range is all $y<-1$ together with all $y\ge 0$, that is $(-\infty,-1)\cup[0,\infty)$. \[ \boxed{(-\infty,-1)\cup[0,\infty)} \]