Let us check for invertibility of $f\left(x\right)$
$\left(A\right)$ one-one: we have, $f\left(x\right)=\frac{e^{2x}-e^{-2x}}{2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{e^{4x}+1}{e^{2x}},$ which is strictly increasing as $e^{4x}>0$ for all $x.$
Thus, $f$ is one-one
$\left(B\right)$ Onto: Let $y=f\left(x\right)$
$\Rightarrow \frac{dy}{dx}=e^{2x}+e^{-2x},$ where y is strictly monotonic
Hence, the range of $f\left(x\right)=\left(f\left(-\mathrm{\infty }\right),f\left(\mathrm{\infty }\right)\right)$
$\Rightarrow$ range of $f\left(x\right)=\left(-\mathrm{\infty },\mathrm{\infty }\right)$
So, the range of $f\left(x\right)=$ co-domain
Hence, $f\left(x\right)$ is one-one and onto
$\left(C\right)$ To find $f^{-1}:y=\frac{e^{4x}-1}{2e^{2x}}$
$\Rightarrow e^{4x}-2e^{2x}y-1=0$
$\Rightarrow e^{2x}=\frac{2y\pm \sqrt{4y^2+4}}{2}$
$\Rightarrow 2x=\mathrm{l}\mathrm{o}\mathrm{g}\left(y\pm \sqrt{y^2+1}\right)$
$\Rightarrow f^{-1}\left(y\right)=\frac{\mathrm{l}\mathrm{o}\mathrm{g}\left(y\pm \sqrt{y^2+1}\right)}{2}$
Since, $e^{f^{-1}\left(x\right)}$ is always positive, so neglecting negative sign.
Hence, $f^{-1}\left(x\right)=\frac{\mathrm{l}\mathrm{o}\mathrm{g}\left(x+\sqrt{x^2+1}\right)}{2}$