$f:R-\left\{0\right\}\to R$ given by $f\left(x\right)=\frac{1}{x}-\frac{3}{e^{3x}-1}$ can be made continuous at $x=0$ by definding $f\left(0\right)$ as

$f\left(0\right)=\underset{x\to 0}{Lt}f\left(x\right)$ 

$$\begin{gathered} f\left(0\right)=\underset{x\to 0}{Lt}\frac{e^{3x}-1-3x}{x(e^{3x}-1)} \\ =\underset{x\to 0}{\lim }\frac{e^{3x}-1-3x}{x^2\left({\displaystyle \frac{e^{3x}-1}{x}}\right)} \end{gathered}$$

         $=\underset{x\to 0}{Lt}\frac{e^{3x}-1-3x}{x^2.3} (\frac{0}{0} form by L\text{'}Hospital rule)$

         $=\underset{x\to 0}{Lt}\frac{e^{3x}.3-3}{2x.3}=\frac{3}{2}$