If $y^{x}=x^{y}$, then $\frac{d y}{d x}$ is :

$\frac{y(x \log y-y)}{x(y \log x-x)}$

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

$\Rightarrow x \log y=y \log x$

$\Rightarrow x . \frac{1}{y} \frac{d y}{d x}+\log y . 1=y . \frac{1}{x}+\log x . \frac{d y}{d x}$

$\Rightarrow\left(\frac{x}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}-\log y$

Hence (2) is correct answer ⇒ $\frac{d y}{d x}=\frac{y(y-x \log y)}{x(x-y \log x)}$