Answer
$a)$ See solution
$b)$ $f^{-1}(x)=\frac{e^{x}-e^{-x}}{2}$
Work Step by Step
$a)$
$$f(-x)=\ln(-x+\sqrt{x^{2}+1})$$
$$f(-x)+f(x)=\ln(-x+\sqrt{x^{2}+1})+\ln(x+\sqrt{x^{2}+1})$$
$$f(-x)+f(x)=\ln((-x+\sqrt{x^{2}+1})(x+\sqrt{x^{2}+1}))$$
$$f(-x)+f(x)=\ln((\sqrt{x^{2}+1})^{2}-x^{2})$$
$$f(-x)+f(x)=\ln(x^{2}+1-x^{2})$$
$$f(-x)+f(x)=\ln(1)$$
$$f(-x)+f(x)=0$$
$$f(-x)=-f(x)$$
so $f$ is an odd function.
$b)$
Solve the following equation for $x$:
$$y=\ln(x+\sqrt{x^{2}+1})$$
$$e^{y}=x+\sqrt{x^{2}+1}$$
also since $f$ is odd it follows that:
$$-y=\ln(-x+\sqrt{x^{2}+1})$$
$$e^{-y}=-x+\sqrt{x^{2}+1}$$
so:
$$e^{y}-e^{-y}=x+\sqrt{x^{2}+1}-(-x+\sqrt{x^{2}+1})$$
$$e^{y}-e^{-y}=2x$$
$$\frac{e^{y}-e^{-y}}{2}=x$$
so the inverse of $f$ is:
$$f^{-1}(x)=\frac{e^{x}-e^{-x}}{2}$$
