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

$\begin{array}{l}\text{ Given }x=e^{y+x}\\ \text{ Take logarithm on both sides }\end{array}$

$\begin{array}{l}\log x=\log e^{y+x}\\ \log x=(y+x)\log e\\ \log x=y+x\end{array}$

Differentiate both sides 

$\frac{1}{x}=\frac{dy}{dx}+1\Rightarrow \frac{dy}{dx}=\frac{1}{x}-1\Rightarrow \frac{dy}{dx}=\frac{1-x}{x}$

$\text{ Therefore, }k=1$