Answer
$n=0,\frac{1}{2}$
Work Step by Step
Given: $\dfrac{n}{n-2}+\dfrac{n}{n+2}=\frac{n}{n^2-4}$
$\dfrac{n}{n-2}+\dfrac{n}{n+2}=\frac{n}{(n-2)(n+2)}$
$\dfrac{n(n+2)}{(n-2)(n+2)}+\dfrac{n(n-2)}{(n-2)(n+2)}=\dfrac{n}{(n-2)(n+2)}$
$n(n+2)+n(n-2)=n$
$n^2+2n+n^2-2n=n$
$n^2+2n+n^2-2n-n=0$
$2n^2-n=0$
$n(2n-1)=0$
Hence, $n=0,\frac{1}{2}$