$\text{ If }x=e^{y+e^{y^{e^{y.....\infty }}}},x=0\text{ then }\frac{dy}{dx}=\frac{1-kx}{x},k=\text{ ? }$

$\begin{array}{l}x=e^{y+x}\\ \log x=\log e^{y+x}\\ \log x=(y+x)\log e\\ \log x=y+x\\ \frac{1}{x}=\frac{dy}{dx}+1\\ \frac{dy}{dx}=\frac{1}{x}-1\\ =\frac{1-x}{x}\therefore k=1\end{array}$