Solution:

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

$u+v=1 \text {, in which } u=x^{y} \text { and } v=y^{z}$

$\frac{d}{d x} u+\frac{d}{d x} v=\frac{d}{d x} 1$

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

So now $u=x^{3}$

$\log u=\log x^{y}=y \log x$

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

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

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

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

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

$\frac{d u}{d x}=x^{y-1} y+x^{y} \log x \cdot \frac{d y}{d x} \dots \dots \dots$(2)

Now again $v=y^{x}$

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

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

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

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

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

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

$\frac{d v}{d x}=y^{x-1} x \frac{d y}{d x}+y^{x} \log y \dots \dots \dots$(3)

Putting values from equations(2) and (3) in equation(1),

$x^{y-1} y+x^{y} \log x \cdot \frac{d y}{d x}+y^{x-1} x \frac{d y}{d x}+y^{x} \log y=0$

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

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