Find of function.

The given function is $y^{x}=x^{y}$'

Taking logarithm on both the sides, we obtain

$x \log y=y \log x$

Differentiating both sides with respect to x, we obtain

$\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x)$

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

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

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

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

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