Example 5 Given $x^{2}+\sqrt{2} y=\sqrt{3}, y^{2}+\sqrt{2} x=$ $\sqrt{3}, x \neq y$. Then the value of $\frac{y}{x}+\frac{x}{y}$ is ().
(A) $2-2 \sqrt{3}$
(B) $2+2 \sqrt{3}$
(C) $2-\sqrt{3}$
(D) $2+\sqrt{3}$
Solution: From $x^{2}+\sqrt{2} y=\sqrt{3}$,
$$
y^{2}+\sqrt{2} x=\sqrt{3} \text {, }
$$
(1) - (2) gives
$$
\begin{array}{l}
\left(x^{2}-y^{2}\right)-\sqrt{2}(x-y)=0 . \\
\because x \neq y, \\
\therefore x+y=\sqrt{2} .
\end{array}
$$
(1) + (2) gives
$$
\left(x^{2}+y^{2}\right)+\sqrt{2}(x+y)=2 \sqrt{3} \text {. }
$$

Thus, $x^{2}+y^{2}=2 \sqrt{3}-2$.
Therefore, $x y=\frac{1}{2}\left[(x+y)^{2}-\left(x^{2}+y^{2}\right)\right]$
$$
=2-\sqrt{3} \text {. }
$$

Hence $\frac{y}{x}+\frac{x}{y}=\frac{x^{2}+y^{2}}{x y}=2+2 \sqrt{3}$.
The answer is (B).