\(y = \log(x + \sqrt{x^2 + 1})\) ......(1)
Diff. equation (1) w.r.t. x
\(\frac{dy}{dx} = \frac{d[\log(x + \sqrt{x^2 + 1})]}{d(x + \sqrt{x^2 + 1})}\times \frac{d[x + \sqrt{x^2 + 1}]}{dx}\) (chain rule)
\(\frac{d(\log x)}{dx} = \frac 1x\)
\(\frac{dy}{dx} = \frac 1{x + \sqrt {x^2 + 1} } \times \left[1 + \frac 1{2\sqrt{x^2 +1}}\times2x\right]\)
\(\frac{dy}{dx} = \frac 1{x + \sqrt{x^2 + 1}} \times \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}\)
\(\frac{dy}{dx} = \frac 1{\sqrt{x^2 + 1}}\)
\(\sqrt{x^2 + 1}. \frac{dy}{dx} = 1\) ......(2)
Diff. equation (2) w.r.t. x
\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{dy}{dx}\times \frac d{dx}(\sqrt {x^2 + 1}) = 0\)
\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{dy}{dx}\times \frac 1{2\sqrt{x^2 + 1}}\times 2x = 0\)
\(\sqrt{x^2 + 1}.\frac{d^2y}{dx^2} + \frac{x.\frac{dy}{dx}}{\sqrt{x^2 + 1}} = 0\)
\(\frac{(x^2 + 1).\frac{d^2y}{dx^2} + x.\frac{dy}{dx}}{\sqrt{x^2 + 1}} = 0\)
\({(x^2 + 1).\frac{d^2y}{dx^2} + x.\frac{dy}{dx}} = 0\)
Hence proved.