Correct Answer: $\frac{2 m n }{x^n(n^2-m^2)}$

Solution : Given, $mx^m − nx^n = 0$
⇒ $mx^m = nx^n$
⇒ $\frac{x^m}{x^n} = \frac{n}{m}$
Now,
$\frac{1}{x^m+x^n}+\frac{1}{x^m-x^n}$
= $\frac{x^m-x^n + x^m+x^n}{(x^m+x^n)(x^m-x^n)}$
= $\frac{2x^m}{(x^m+x^n)(x^{2m}-x^n)}$
= $\frac{n-m+n+m}{(n+m)(n-m)}$
= $\frac{2n}{(n^2−m^2)}$
Now, the value of $\frac{1}{x^m+x^n}+\frac{1}{x^m−x^n}$ in terms of $x^n$
= $\frac{2n}{n^2-m^2}\times\frac{x^n}{x^n}$
= $\frac{2nm}{x^n(n^2-m^2)}$ [As $x^n=m$]
Hence, the correct answer is $\frac{2mn}{x^n(n^2-m^2)}$.