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

$x^y=y^x$

$\log x^y=\log y^x$

$y\log x=x\log y$

Differentiating

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

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

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

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