Answer
$\dfrac{4x}{x^2+4}$
Work Step by Step
Let's note:
$$E=\dfrac{\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}}{\dfrac{x-2}{x+2}+\dfrac{x+2}{x-2}}.$$
Multiply both numerator and denominator by $(x-2)(x+2)$ and simplify:
$$\begin{align*}
E&=\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\cdot \dfrac{\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}}{\dfrac{x-2}{x+2}+\dfrac{x+2}{x-2}}\\
&=\dfrac{(x-2)(x+2)\dfrac{x+2}{x-2}-(x-2)(x+2)\dfrac{x-2}{x+2}}{(x-2)(x+2)\dfrac{x-2}{x+2}+(x-2)(x+2)\dfrac{x+2}{x-2}}\\
&=\dfrac{(x+2)^2-(x-2)^2}{(x-2)^2+(x+2)^2}\\
&=\dfrac{x^2+4x+4-x^2+4x-4}{x^2-4x+4+x^2+4x+4}\\
&=\dfrac{8x}{2x^2+8}\\
&=\dfrac{8x}{2(x^2+4)}\\
&=\dfrac{4x}{x^2+4}.
\end{align*}$$
