Find of function.

The given function is $x^{y}+y^{x}=1$

Let $x^{y}=u$ and $y^{x}=v$

Then, the function becomes u + v = 1

$\therefore \frac{d u}{d x}+\frac{d v}{d x}=0$  ...(1)

$u=x^{y}$

$\Rightarrow \log u=\log \left(x^{y}\right)$

$\Rightarrow \log u=y \log x$

Differentiating both sides with respect to x, we obtain

$\frac{1}{u} \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^{y}\left(\log x \frac{d y}{d x}+\frac{y}{x}\right)$   ...(2)

$v=y^{x}$

$\Rightarrow \log v=\log \left(y^{x}\right)$

$\Rightarrow \log v=x \log y$

Differentiating both sides with respect to x, we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)$

$\Rightarrow \frac{d v}{d x}=v\left(\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right)$

$\Rightarrow \frac{d v}{d x}=y^{x}\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)$   ...(3)

From (1), (2), and (3), we obtain

$x^{y}\left(\log x \frac{d y}{d x}+\frac{y}{x}\right)+y^{x}\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)=0$

$\Rightarrow\left(x^{y} \log x+x y^{x-1}\right) \frac{d y}{d x}=-\left(y x^{y-1}+y^{x} \log y\right)$

$\therefore \frac{d y}{d x}=-\frac{y x^{y-1}+y^{x} \log y}{x^{y} \log x+x y^{x-1}}$