$z$ is a complex number and $|z|=1$ and $z^2\ne1$. Then $\frac z{1-z^2}$ lies on
$\textbf{(A)}~\text{a line not through origin}$
$\textbf{(B)}~\text{|z|=2}$
$\textbf{(C)}~x-\text{axis}$
$\textbf{(D)}~y-\text{axis}$

Given that \( z \) is a complex number with \( |z| = 1 \) and \( z^2 \neq 1 \), we need to determine the geometric location of \( \frac{z}{1 - z^2} \).

1. **Express \( z \) in polar form:**
   Since \( |z| = 1 \), we can write \( z \) as \( z = e^{i\theta} \) for some real number \( \theta \).

2. **Calculate \( z^2 \):**
   \[
   z^2 = (e^{i\theta})^2 = e^{i2\theta}
   \]

3. **Calculate \( 1 - z^2 \):**
   \[
   1 - z^2 = 1 - e^{i2\theta}
   \]

4. **Find the modulus of \( 1 - z^2 \):**
   \[
   |1 - z^2| = |1 - e^{i2\theta}|
   \]
   Since \( e^{i2\theta} \) lies on the unit circle, the distance from \( 1 \) to \( e^{i2\theta} \) is the chord length of the unit circle subtended by the angle \( 2\theta \). Using the formula for the chord length:
   \[
   |1 - e^{i2\theta}| = 2 \sin(\theta)
   \]

5. **Calculate \( \frac{z}{1 - z^2} \):**
   \[
   \frac{z}{1 - z^2} = \frac{e^{i\theta}}{1 - e^{i2\theta}}
   \]

6. **Simplify the expression:**
   \[
   \frac{e^{i\theta}}{1 - e^{i2\theta}} = \frac{e^{i\theta}}{2i \sin(\theta)} = \frac{e^{i\theta}}{2i \sin(\theta)} = \frac{e^{i\theta}}{2i \sin(\theta)}
   \]
   Since \( e^{i\theta} = \cos(\theta) + i \sin(\theta) \):
   \[
   \frac{\cos(\theta) + i \sin(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)} + \frac{i \sin(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)} + \frac{1}{2}
   \]
   Simplifying further:
   \[
   \frac{\cos(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)}
   \]
   Since \( \cos(\theta) = \cos(\theta) \):
   \[
   \frac{\cos(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)} = \frac{\cos(\theta)}{2i \sin(\theta)}
   \]

7. **Determine the geometric location:**
   The imaginary part of \( \frac{z}{1 - z^2} \) is zero, which means it lies on the y-axis.

The final answer is \(\boxed{\text{y-axis}}\).