Answer
$$
f(x)=\ln (x+\sqrt{x^{2}+1})
$$
(a)
$$
\begin{aligned} f(-x) &=\ln (-x+\sqrt{(-x)^{2}+1}) \\
&=\ln \left(-x+\sqrt{x^{2}+1} \cdot \frac{-x-\sqrt{x^{2}+1}}{-x-\sqrt{x^{2}+1}}\right) \\ &=\ln \left(\frac{x^{2}-\left(x^{2}+1\right)}{-x-\sqrt{x^{2}+1}}\right)=\ln \left(\frac{-1}{-x-\sqrt{x^{2}+1}}\right)\\
&=\ln \left(\frac{1}{x+\sqrt{x^{2}+1}}\right) \\
&=\ln 1-\ln (x+\sqrt{x^{2}+1}) \\
&=-\ln (x+\sqrt{x^{2}-1}) \\
&=-f(x) \end{aligned}
$$
Therefore, the given function is an odd function.
(b)
The inverse function of $f$ is
$$
f^{-1}(x) =\frac{e^{2 x}-1}{2 e^{x}}.
$$
Work Step by Step
$$
f(x)=\ln (x+\sqrt{x^{2}+1})
$$
(a)
$$
\begin{aligned} f(-x) &=\ln (-x+\sqrt{(-x)^{2}+1}) \\
&=\ln \left(-x+\sqrt{x^{2}+1} \cdot \frac{-x-\sqrt{x^{2}+1}}{-x-\sqrt{x^{2}+1}}\right) \\ &=\ln \left(\frac{x^{2}-\left(x^{2}+1\right)}{-x-\sqrt{x^{2}+1}}\right)=\ln \left(\frac{-1}{-x-\sqrt{x^{2}+1}}\right)\\
&=\ln \left(\frac{1}{x+\sqrt{x^{2}+1}}\right) \\
&=\ln 1-\ln (x+\sqrt{x^{2}+1}) \\
&=-\ln (x+\sqrt{x^{2}-1}) \\
&=-f(x) \end{aligned}
$$
Therefore, the given function is an odd function.
(b)
$$
y=\ln (x+\sqrt{x^{2}+1})
$$
Interchanging $x $ and $y $, we get
$$
x=\ln (y+\sqrt{y^{2}+1}) \Rightarrow e^{x}=y+\sqrt{y^{2}+1}
$$
$ \Rightarrow$
$$
e^{x}-y=\sqrt{y^{2}+1} \Rightarrow e^{2 x}-2 y e^{x}+y^{2}=y^{2}+1 \Rightarrow e^{2 x}-1=2 y e^{x} \Rightarrow y=\frac{e^{2 x}-1}{2 e^{x}}=f^{-1}(x).
$$
