Example 6 Let $x_{1}=1, x_{n+1}=\frac{x_{n}^{2}}{\sqrt{3 x_{n}^{4}+6 x_{n}^{2}+2}}$ $(n \geqslant 1)$. Find $x_{n}$.
Given the original recurrence relation, squaring both sides yields
$$
x_{n+1}^{2}=\frac{x_{n}^{4}}{3 x_{n}^{4}+6 x_{n}^{2}+2} \text {. }
$$

Let $y_{n}=x_{n}^{2}$. Then $y_{n+1}=x_{n+1}^{2}$.
Thus, $y_{n+1}=\frac{y_{n}^{2}}{3 y_{n}^{2}+6 y_{n}+2}$.
Therefore, $\frac{1}{y_{n+1}}=\frac{3 y_{n}^{2}+6 y_{n}+2}{y_{n}^{2}}$ $=3+6 \cdot \frac{1}{y_{n}}+2\left(\frac{1}{y_{n}}\right)^{2}$.
Let $t_{n}=\frac{1}{y_{n}}$. Then $t_{n+1}=\frac{1}{y_{n+1}}$.
Hence, $t_{n+1}=3+6 t_{n}+2 t_{n}^{2}$
$$
=2\left(t_{n}+\frac{3}{2}\right)^{2}-\frac{3}{2} \text {. }
$$

Thus, $t_{n+1}+\frac{3}{2}=2\left(t_{n}+\frac{3}{2}\right)^{2}$.
Taking the common logarithm of both sides, we get
$$
\lg \left(t_{n+1}+\frac{3}{2}\right)=\lg 2+2 \lg \left(t_{n}+\frac{3}{2}\right) \text {. }
$$

Let $b_{n}=\lg \left(t_{n}+\frac{3}{2}\right)$, then
$$
b_{n+1}=\lg \left(t_{n+1}+\frac{3}{2}\right) \text {. }
$$

Thus, $b_{n+1}=\lg 2+2 b_{n}$, which means
$$
b_{n+1}+\lg 2=2\left(b_{n}+\lg 2\right) \text {. }
$$

Therefore, $\frac{b_{n+1}+\lg 2}{b_{n}+\lg 2}=2$.
Given $x_{1}=1$, we have $y_{1}=1$, and thus $t_{1}=1$.
Hence, $b_{1}=\lg \left(1+\frac{3}{2}\right)=\lg \frac{5}{2}$, which means
$$
b_{1}+\lg 2=\lg \frac{5}{2}+\lg 2=\lg 5 \text {. }
$$

Therefore, $b_{n}+\lg 2=(\lg 5) 2^{n-1}$, which means
$$
\lg \left(t_{n}+\frac{3}{2}\right)+\lg 2=\lg 5^{2^{n-1}} \text {. }
$$

Thus, $2\left(t_{n}+\frac{3}{2}\right)=5^{2^{n-1}}$, which means
$$
2 t_{n}+3=5^{2^{n-1}} \text {. }
$$

Hence, $\frac{2}{y_{n}}+3=5^{2^{n-1}}$, which means $\frac{2}{x_{n}^{2}}+3=5^{2^{n-1}}$.
Since $x_{n}>0$, we have $x_{n}=\frac{\sqrt{2}}{\sqrt{5^{2^{n-1}}-3}}$.