$\begin{align}  & y=\log \left( x+\sqrt{1+{{x}^{2}}} \right) \\ & y=\log u;\text{ }\frac{dy}{du}=\frac{1}{u} \\ & u=x+\sqrt{1+{{x}^{2}}},\text{  }\frac{du}{dx}=1+\frac{2x}{2\sqrt{1+{{x}^{2}}}}=1+\frac{x}{\sqrt{1+{{x}^{2}}}} \\ & \frac{du}{dx}=\frac{\sqrt{1+{{x}^{2}}}+x}{\sqrt{1+{{x}^{2}}}} \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=\frac{1}{u}\times \frac{\sqrt{1+{{x}^{2}}}+x}{\sqrt{1+{{x}^{2}}}} \\ & \frac{dy}{dx}=\frac{1}{x+\sqrt{1+{{x}^{2}}}}\times \frac{\sqrt{1+{{x}^{2}}}+x}{\sqrt{1+{{x}^{2}}}} \\ & \frac{dy}{dx}=\frac{1}{\sqrt{1+{{x}^{2}}}} \\ & \sqrt{1+{{x}^{2}}}\frac{dy}{dx}=1 \\ & \frac{d}{dx}\left( \sqrt{1+{{x}^{2}}}\frac{dy}{dx} \right)=\frac{d}{dx}(1) \\ & \sqrt{1+{{x}^{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\times \left( \frac{1}{2\sqrt{1+{{x}^{2}}}}\times 2x \right)=0 \\ & \sqrt{1+{{x}^{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\times \left( \frac{x}{\sqrt{1+{{x}^{2}}}} \right)=0 \\ & \sqrt{1+{{x}^{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{x}{\sqrt{1+{{x}^{2}}}}\frac{dy}{dx}=0 \\ & \text{Multiply both sides by }\sqrt{1+{{x}^{2}}} \\ & (1+{{x}^{2}})+x\frac{dy}{dx}=0 \\\end{align}$