If $f(x)=\left\{\begin{array}{cc}\frac{1}{x}-\frac{2}{e^{2 x}-1}, & x \neq 0 \\ 1, & x=0\end{array}\right.$, then

f(x) is differentiable at x = 0

We have,

$\lim\limits_{x \rightarrow 0} f(x) =\lim\limits_{x \rightarrow 0}\left(\frac{1}{x}-\frac{2}{e^{2 x}-1}\right)$

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

$\Rightarrow \lim\limits_{x \rightarrow 0} f(x) =\lim\limits_{x \rightarrow 0} \frac{\left(1+2 x+\frac{(2 x)^2}{2 !}+\frac{(2 x)^3}{3 !}+...\right)-1-2 x}{x\left(1+2 x+\frac{(2 x)^2}{2 !}+... .-1\right)}$

$\Rightarrow \lim\limits_{x \rightarrow 0} f(x)=\lim\limits_{x \rightarrow 0} \frac{\frac{(2 x)^2}{2 !}+\frac{(2 x)^3}{3 !}+...}{x\left(2 x+\frac{(2 x)^2}{2 !}+...\right)}$

$\Rightarrow \lim\limits_{x \rightarrow 0} f(x)=\frac{\frac{2^2}{2 !}}{2}=1=f(0)$

So, f(x) is continuous at x = 0.

Now,

(LHD at x = 0) = $\lim\limits_{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$

⇒ (LHD at x = 0) = $\lim\limits_{h \rightarrow 0} \frac{f(-h)-1}{-h}$

⇒ (LHD at x = 0) = $\lim\limits_{h \rightarrow 0} \frac{\left(-\frac{1}{h}-\frac{2}{e^{-2 h}-1}-1\right)}{-h}$

⇒ (LHD at x = 0) = $\lim\limits_{h \rightarrow 0} \frac{e^{-2 h}-1+2 h+h\left(e^{-2 h}-1\right)}{h^2\left(e^{-2 h}-1\right)} \left\{1-2 h+\frac{(2 h)^2}{2 !}-\frac{(2 h)^3}{3 !}+...\right\}$

$=\lim\limits_{h \rightarrow 0} \frac{+h\left\{-2 h+\frac{(2 h)^2}{2 !}-\frac{(2 h)^3}{3 !}+...\right\}-1+2 h}{h^2\left(-2 h+\frac{(2 h)^2}{2 !}-\frac{(2 h)^3}{3 !}+...\right)}$

$=\lim\limits_{h \rightarrow 0} \frac{\left(\frac{-8}{3 !}+\frac{4}{2 !}\right) h^3+...}{h^3\left(-2+\frac{4 h}{2 !}-...\right)}=-\frac{1}{3}$

Similarly, we have

(RHD at x = 0) = $-\frac{1}{3}$

∴ (LHD at x = 0) = (RHD at x = 0)

Thus, f(x) is differentiable at x = 0.