Given $f\left(x\right)=\frac{1}{x}-\frac{k-1}{e^{2x}-1}, x\ne 0$

and $f\left(x\right)$ is continuous at $x=0$.

$\Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{k-1}{e^{2x}-1}\right)$

$=\underset{x\to 0}{\lim }\frac{\left(1+\left(2x\right)+\frac{1}{2!}{\left(2x\right)}^2+\dots \left(-1-x\left(k-1\right)\right)\right)}{2x^2\left(\frac{e^{2x}-1}{2x}\right)}$

$=\underset{x\to 0}{\lim }\frac{\left(1+\left(2x\right)+\frac{1}{2!}{\left(2x\right)}^2+\dots \left(-1-x\left(k-1\right)\right)\right)}{2x^2}$

Clearly, $k=3$ and $f\left(0\right)=1$. $($ $\because$Equating coefficient of $x=0$$)$