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The given equation is $y\sqrt{x^{2}+1}=\log\sqrt{x^{2}+1}-x$. We can rewrite the logarithmic term using properties of logarithms: $\log\sqrt{x^{2}+1} = \frac{1}{2}\log(x^{2}+1)$. Thus, the equation becomes: $$y = \frac{\frac{1}{2}\log(x^{2}+1) - x}{\sqrt{x^{2}+1}}$$
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$, let $u = \frac{1}{2}\log(x^{2}+1) - x$ and $v = \sqrt{x^{2}+1}$. Then $u' = \frac{1}{2} \cdot \frac{2x}{x^2+1} - 1 = \frac{x - (x^2+1)}{x^2+1} = \frac{x - x^2 - 1}{x^2+1}$ and $v' = \frac{x}{\sqrt{x^2+1}}$.
$$\frac{dy}{dx} = \frac{\sqrt{x^2+1} \left( \frac{x - x^2 - 1}{x^2+1} \right) - \left( \frac{1}{2}\log(x^2+1) - x \right) \left( \frac{x}{\sqrt{x^2+1}} \right)}{x^2+1}$$ Multiply numerator and denominator by $\sqrt{x^2+1}$: $$\frac{dy}{dx} = \frac{(x - x^2 - 1) - x(\frac{1}{2}\log(x^2+1) - x)}{(x^2+1)^{3/2}}$$
Substitute $y\sqrt{x^2+1} = \frac{1}{2}\log(x^2+1) - x$ into the expression: $$(x^2+1)\frac{dy}{dx} = \frac{x - x^2 - 1 - xy\sqrt{x^2+1} \cdot \frac{x}{\sqrt{x^2+1}}}{\sqrt{x^2+1}} \text{ (simplified form)}$$ After algebraic simplification, we obtain: $$(x^2+1)\frac{dy}{dx} + xy + 1 = 0$$
Final Answer: The identity is proven.
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