4. Suppose $x, y$ are real numbers such that $\frac{1}{x}-\frac{1}{2 y}=\frac{1}{2 x+y}$. Find the value of $\frac{y^{2}}{x^{2}}+\frac{x^{2}}{y^{2}}$.
(A) $\frac{2}{3}$
(B) $\frac{9}{2}$
(C) $\frac{9}{4}$
(D) $\frac{4}{9}$
(E) $\frac{2}{9}$
4. Answer: (C)
$$
\begin{aligned}
\frac{1}{x}-\frac{1}{2 y}=\frac{1}{2 x+y} & \Rightarrow \frac{2 x+y}{x}-\frac{2 x+y}{2 y}=1 \\
& \Rightarrow 2+\frac{y}{x}-\frac{x}{y}-\frac{1}{2}=1 \\
& \Rightarrow \frac{y}{x}-\frac{x}{y}=-\frac{1}{2}
\end{aligned}
$$

Now we have
$$
\frac{y^{2}}{x^{2}}+\frac{x^{2}}{y^{2}}=\left(\frac{y}{x}-\frac{x}{y}\right)^{2}+2=\left(-\frac{1}{2}\right)^{2}+2=\frac{9}{4} \text {. }
$$