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

$\mathrm{e}^{\log x^{x}}+\mathrm{e}^{\log y^{x}}=1$

$\mathrm{e}^{\mathrm{xlogx}}+\mathrm{e}^{\mathrm{xlog} y}=1$

$\left[\right.$ Since $\left.\mathrm{e}^{\log a}=\mathrm{a} \cdot \log \mathrm{a}^{\mathrm{b}}=\mathrm{b} \log \mathrm{a}\right]$

Differentiating it with respect to $x$ using chain rule and product rule,

$\frac{d}{d x} e^{x \log x}+\frac{d}{d x} e^{x \log y}=\frac{d}{d x}(1)$

$e^{x \log x} \frac{d}{d x}(x \log x)+e^{x \log y} \frac{d}{d x}(x \operatorname{logy})=0$

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

$x^{x}\left[x\left(\frac{1}{x}\right)+\log x(1)\right]+y^{x}\left[x\left(\frac{1}{y}\right)+\log y(1)\right]=0$

$x^{x}[1+\log x]+y^{x}\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0$

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

$\left(x y^{x-1}\right) \frac{d y}{d x}=-\left[x^{x}(1+\log x)+y^{x} \log y\right]=0$

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

Hence Proved.