$\newcommand\ep{\epsilon}$$\newcommand\de{\delta}$Note that $l(x):=(\ln x)/x$ is decreasing in $x\ge e$. For $\ep\in(0,1/e]$, letting 
$$y:=y_\ep:=\frac1\ep\,\ln\frac1\ep\ge e,$$
we have 
$$l(y)=\ep\frac{\ln\frac1\ep+\ln\ln\frac1\ep}{\ln\frac1\ep}\ge\ep$$
and hence 
$$x_\ep\ge y=\frac1\ep\,\ln\frac1\ep,$$
where $x_\ep\in[e,\infty)$ is the root of the equation
$$l(x_\ep)=\ep.$$
On the other hand, for each real $\de>0$, letting 
$$z:=z_\ep:=(1+\de)y_\ep$$
we have 
$$l(y)=\ep\frac{\ln\frac{1+\de}\ep+\ln\ln\frac1\ep}{(1+\de)\ln\frac1\ep}\le\ep$$
for all small enough $\ep>0$ and hence 
$$x_\ep\le z=\frac{1+\de}\ep\,\ln\frac1\ep.$$