Four, (16 points) Function
$$
y=\left|x^{2}-1\right|+\left|2 x^{2}-1\right|+\left|3 x^{2}-1\right|
$$

Find the range of the independent variable $x$ when the function achieves its minimum value of 1.
Let $x^{2}=t(t \geqslant 0)$. Then
$$
y=|t-1|+|2 t-1|+|3 t-1| \text {. }
$$

When $0 \leqslant t < 1$,
$$
\begin{array}{l}
y=t-1+2 t-1+3 t-1=6 t-3 \\
\Rightarrow y>3 .
\end{array}
$$

In summary, when $\frac{1}{3} \leqslant t \leqslant \frac{1}{2}$, $y_{\text {min }}=1$.
Thus, from $\frac{1}{3} \leqslant x^{2} \leqslant \frac{1}{2}$, when $y_{\min }=1$,
$$
-\frac{\sqrt{2}}{2} \leqslant x \leqslant-\frac{\sqrt{3}}{3} \text { or } \frac{\sqrt{3}}{3} \leqslant x \leqslant \frac{\sqrt{2}}{2} \text {. }
$$