For continuity at $x=0$, we must have $f(0)=\lim_{x\to 0}\left(\frac{1}{x}-\frac{k-1}{e^{2x}-1}\right).$ So first we evaluate this limit and ensure it is finite.

Use the expansion of $e^{2x}$ near $x=0$: $e^{2x}=1+2x+2x^2+\cdots$ Hence, $e^{2x}-1=2x+2x^2+\cdots = 2x(1+x+\cdots).$ Therefore, $\frac{1}{e^{2x}-1}\sim \frac{1}{2x}\quad (x\to 0).$ So $\frac{k-1}{e^{2x}-1}\sim \frac{k-1}{2x}.$

Then near $x=0$, $f(x)=\frac{1}{x}-\frac{k-1}{e^{2x}-1}\sim \frac{1}{x}-\frac{k-1}{2x}.$ For the limit to exist finitely, the coefficient of $\frac{1}{x}$ must vanish: $1-\frac{k-1}{2}=0.$ Solve: $2-(k-1)=0 \Rightarrow 3-k=0 \Rightarrow k=3.$

Now substitute $k=3$: $f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}.$ We need $f(0)=\lim_{x\to 0}\left(\frac{1}{x}-\frac{2}{e^{2x}-1}\right).$ Rewrite as $f(x)=\frac{e^{2x}-1-2x}{x(e^{2x}-1)}.$

Use expansion again: $e^{2x}-1=2x+2x^2+\frac{4}{3}x^3+\cdots$ So $e^{2x}-1-2x=2x^2+\frac{4}{3}x^3+\cdots$ Also, $x(e^{2x}-1)=x\left(2x+2x^2+\cdots\right)=2x^2+2x^3+\cdots$ Thus, $\lim_{x\to 0}\frac{e^{2x}-1-2x}{x(e^{2x}-1)}=\frac{2}{2}=1.$ Therefore, $f(0)=1.$

Hence the ordered pair is $(k,f(0))=(3,1).$

Checking options: