$f\left(x\right)=\frac{a^x-1}{x^n\left(a^x+1\right)}$

since, $f\left(x\right)$ is symmetrical about $Y$-axis.

( i.e., it is a even function.)

$\therefore f\left(x\right)=f\left(-x\right)$

$\Rightarrow \frac{a^x-1}{x^n\left(a^x+1\right)}=\frac{a^{-x}-1}{{\left(-x\right)}^n\left(a^{-x}+1\right)}$

$\Rightarrow \frac{a^x-1}{x^n\left(a^x+1\right)}=\frac{1-a^x}{{\left(-x\right)}^n\left(1+a^x\right)}$

$\Rightarrow x^n=-{\left(-x\right)}^n$

Hence, the value of $n$ which satisfy this relation, is $-\frac{1}{3}$