If the function f defined as $f\left(x\right)=\frac{1}{x}-\frac{k-1}{e^{2x}-1},$$x\ne 0$ is continuous at $x=0$ then the ordered  pair $\left(k,f\left(0\right)\right)$ is

$\underset{x\to 0}{Lt}f\left(x\right)=\underset{x\to 0}{Lt}\frac{e^{2x}-1-kx+x}{x\left(e^{2x}-1\right)}$

$\begin{array}{l}=\underset{x\to 0}{Lt}\frac{2e^{2x}-0-k+1}{x\left(2e^{2x}\right)+\left(e^{2x}-1\right)\left(1\right)}\\ =\underset{x\to 0}{Lt}\frac{4e^{2x}}{x\left(4e^{2x}\right)+\left(2e^{2x}\right)+2e^{2x}}\\ =\frac{4}{2+2}=1=f\left(0\right)\\ Also2-k+1=0\Rightarrow k=3.\end{array}$