If $g\left(x\right)$ is a curve which is obtained by the reflection of $f\left(x\right)=\frac{e^x-e^{-x}}{2}$ by the line $y=x$, then

 As $g\left(x\right)$ is a curve, obtained by the reflection of

$f\left(x\right)\frac{e^x-e^{-x}}{2}$ on $y=x$

so $g\left(x\right)$ is inverse of $f\left(x\right)$

$\therefore g\left(x\right)=\log \left(x\sqrt{1x^2}\right)=f^{-1}\left(x\right)$

$$\begin{gathered} \Rightarrow g^{\text{'}}\left(x\right)=\frac{1}{x\sqrt{1x^2}}·\left(1\frac{2x}{2\sqrt{1x^2}}\right) \\ =\frac{1}{\sqrt{1x^2}}\ne 0,\forall x\in R \end{gathered}$$

$\Rightarrow g\left(x\right)$ has no tangent parallel to $X$-axis. Also, $g^{\text{'}}\left(x\right)$ is always defined, $\forall x\in R$.

$\Rightarrow g\left(x\right)$ has no tangent parallel to $Y$-axis.

Since $g^{\text{'}}\left(x\right)>0\forall x\in R$, therefore $g\left(x\right)$ doesn't have any extremum.

Hence, $4$ is the correct answer.