Step 1:
$y=\log(x+\sqrt{x^2+1})$
$\Rightarrow \large\frac{dy}{dx}=\bigg[\large\frac{1}{x+\sqrt{x^2+1}}.$$(1+\large\frac{1}{2\sqrt{x^2+1}.2x}\bigg]$
$\Rightarrow \large\frac{1}{x+\sqrt{x^2+1}}\bigg[\large\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\bigg]$
$\Rightarrow \large\frac{1}{\sqrt{x^2+1}}$
Step 2:
$\large\frac{d^2y}{dx^2}=\large\frac{0-1.1/2\sqrt{x^2+1}.2x}{(\sqrt{x^2+1)^2}}$
$\Rightarrow \large\frac{-x}{(x^2+1)\sqrt{x^2+1}}$
$\Rightarrow (x^2+1)\large\frac{d^2y}{dx^2}=\frac{-x}{\sqrt{x^2+1}}$
$\therefore (x^2+1)\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}=\frac{-x}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}$
$\Rightarrow 0$
Hence proved.