If $e^y = \log x$, then which of the following is true?

Given $e^y = \log x$

Differentiating both sides with respect to $x$:

$e^y \cdot \frac{dy}{dx} = \frac{1}{x}$

$\frac{dy}{dx} = \frac{1}{x \cdot e^y}$

Since $e^y = \log x$:

$\frac{dy}{dx} = \frac{1}{x \log x}$

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Differentiating $\frac{dy}{dx} = \frac{1}{x \log x}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}[(x \log x)^{-1}]$

$= -(x \log x)^{-2} \cdot \frac{d}{dx}(x \log x)$

$\frac{d}{dx}(x \log x) = \log x + x \cdot \frac{1}{x}$

$= \log x + 1$

Therefore:

$\frac{d^2y}{dx^2} = -\frac{\log x + 1}{(x \log x)^2}$

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Checking if $x\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0$:

$x \cdot \left(-\frac{\log x + 1}{(x \log x)^2}\right) + x \cdot \left(\frac{1}{x \log x}\right)^2 + \frac{1}{x \log x}$

$= -\frac{\log x + 1}{x(\log x)^2} + \frac{1}{x(\log x)^2} + \frac{\log x}{x(\log x)^2}$

$= \frac{-\log x - 1 + 1 + \log x}{x(\log x)^2}$

$= \frac{0}{x(\log x)^2}$

$= 0$

Therefore, $x\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0$ is true.