4・118 Solve the system of equations
$$\left\{\begin{array}{l}
\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}=\frac{7}{\sqrt{x y}} \\
\sqrt[4]{x^{3} y}-\sqrt[4]{x y^{3}}=\sqrt{12}
\end{array}\right.$$
[Solution] For $x>0$, then $y>0$. From (1) we get $x-y=7$,
(3)
From (2) we get $\sqrt[4]{x y}(\sqrt{x}-\sqrt{y})=\sqrt{12}$,
squaring both sides, we get $\sqrt{x y}(\sqrt{x}-\sqrt{y})^{2}=12$.
From (3) we get, $\sqrt{x}-\sqrt{y}=\frac{7}{\sqrt{x}+\sqrt{y}}$,
substituting into (4) and simplifying, we get
$$12 x+12 y-25 \sqrt{x y}=0$$

Solving it, we get $\quad \frac{\sqrt{x}}{\sqrt{y}}=\frac{4}{3}, \quad \frac{x}{y}=\frac{16}{9}$.
or $\frac{\sqrt{x}}{\sqrt{y}}=\frac{3}{4}, \frac{x}{y}=\frac{9}{16}$.
Combining with (3), we get
$$\left\{\begin{array}{l}
x_{1}=16 \\
y_{1}=9
\end{array}\right.$$

Upon verification, the solution to the original system of equations is $\left\{\begin{array}{l}x=16, \\ y=9 .\end{array}\right.$
Similarly, for $x<0, y<0$, we get another solution $\left\{\begin{array}{l}x=-16, \\ y=-9 .\end{array}\right.$