If $x = {e^{x/y}},$ then prove that $\frac{{dy}}{{dx}} = \frac{{x - y}}{{x\log x}}$

We have, $x = {e^{x/y}}$
therefore,$\frac{d}{{dx}}x = \frac{d}{{dx}}{e^{x/y}}$

$\Rightarrow$ $1 = {e^{x/y}} \cdot \frac{d}{{dx}}(x/y)$

$\Rightarrow$ $1 = {e^{x/y}} \cdot \left[ {\frac{{y \cdot 1 - x \cdot dy/dx}}{{{y^2}}}} \right]$

$\Rightarrow$ ${y^2} = y \cdot {e^{x/y}} - x \cdot \frac{{dy}}{{dx}} \cdot {e^{x/y}}$

$\Rightarrow$ $x \cdot \frac{{dy}}{{dx}} \cdot {e^{x/y}} = y{e^{x/y}} - {y^2}$

therefore,$\frac{{dy}}{{dx}} = \frac{{y\left( {{e^{x/y}} - y} \right)}}{{x \cdot {e^{x/y}}}}$

$= \frac{{\left( {{e^{x/y}} - y} \right)}}{{{e^{x/y}} \cdot \frac{x}{y}}}$

$= \frac{{x - y}}{{x \cdot \log x}}$