If $x^y=y^x\text{ then show that }\frac{dy}{dx}=\frac{y(x\log y-y)}{x(y\log x-x)}$

Given $x^y=y^x$

Apply log on both sides

$y\log x=x\log y$

Diff. w.r.to $x$

$y1/x+\log x\cdot \frac{dy}{dx}=x\cdot \frac{1}{y}\cdot \frac{dy}{dx}+\log y$

$\frac{y}{x}+\log x\cdot \frac{dy}{dx}=\frac{x}{y}\cdot \frac{dy}{dx}+\log y$

$\frac{dy}{dx}(\log x-x/y)=\log y-y/x$

$\frac{dy}{dx}\left(\frac{y\log x-x}{y}\right)=\left(\frac{x\log y-y}{x}\right)$

$\frac{dy}{dx}=\frac{y(x\log y-y)}{x(y\log x-x)}$