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

Given that $x^y=y^x$

Use logarithmic differentiation, apply logarithm both sides

$y\log x=x\log y$

Differentiate both sides with respect to $x$

$y\left(\frac{1}{x}\right)+\log x\left(\frac{dy}{dx}\right)=x\left(\frac{1}{y}\frac{dy}{dx}\right)+\log y$

separate the $\frac{dy}{dx}$ terms

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

simplify

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

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