If $\mathrm{x}=\mathrm{e}^{\mathrm{x} / \mathrm{y}}$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x} \log \mathrm{x}}$
$x=e^{x} / y$
Taking logon both sides,
$\log x=\log e^{x} / y$
$\log x=\frac{x}{y} \ldots \ldots$ (i) [ Since $\log e^{a}=a$ ]
or, $y=\frac{x}{\log x}$ (ii)
Differentiating the given equation with respect to $x$,
$\frac{d y}{d x}=\frac{\log x \frac{d}{d x}(x)-x \frac{d}{d x}(\log x)}{(\log x)^{2}}$
$\frac{d y}{d x}=\frac{\log x-x \times \frac{1}{x}}{(\log x)^{2}}$
$\frac{d y}{d x}=\frac{\log x-1}{(\log x)^{2}}$
$\frac{d y}{d x}=\frac{\frac{x}{y}-1}{(\log x)^{2}}[$ From (i) $]$
$\frac{d y}{d x}=\frac{x-y}{y(\log x)^{2}}$
$\frac{d y}{d x}=\frac{x-y}{\frac{x}{\log x}(\log x)^{2}}[$ From (ii) $]$
Therefore, $\frac{d y}{d x}=\frac{x-y}{x \log x}$