${x^y} + {y^x} = 1$

${x^y} + {y^x} = 1$ …(i)

Differentiating (i) w.r.t. x, we get

$\cfrac{d}{{dx}}({x^y}) + \cfrac{d}{{dx}}({y^x}) = 0$ …(ii)

Let $u = {x^y}$
therefore, $\log u = y\log x$

Differentiating w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{u}\cfrac{{du}}{{dx}} = y \cdot \cfrac{1}{x} + \log x\cfrac{{dy}}{{dx}}$

$\Rightarrow$ $\cfrac{{du}}{{dx}} = {x^y}\left[ {\cfrac{y}{x} + \log x\cfrac{{dy}}{{dx}}} \right]$ ….(iii)

Let $v = {y^x} \Rightarrow \log v = x\log y$

Differentiating w.r.t. x,we get

$\cfrac{1}{v} \cdot \cfrac{{dv}}{{dx}} = x \cdot \cfrac{1}{y} \cdot \cfrac{{dy}}{{dx}} + \log y$
therefore, $\cfrac{{dv}}{{dx}} = {y^x}\left( {\cfrac{x}{y}\cfrac{{dy}}{{dx}} + \log y} \right)$ ….(iv)

Substituting the values of (iii) and (iv) in (ii), we get

${x^y}\left( {\cfrac{y}{x} + \log x\cfrac{{dy}}{{dx}}} \right) + {y^x}\left( {\cfrac{x}{y}\cfrac{{dy}}{{dx}} + \log y} \right) = 0$

$\Rightarrow$ $({x^y}\log x + x{y^{x - 1}})\cfrac{{dy}}{{dx}} = - ({y^x}\log y + y{x^{y - 1}})$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = - \left( {\cfrac{{{y^x}\log y + y{x^{y - 1}}}}{{{x^y}\log x + x{y^{x - 1}}}}} \right)$