calculus - indeterminate limit where applying L'Hopitals Rules directly doesn't help and using ln gives wrong answer

I am trying to determine the limit $\displaystyle{ \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}}}$. Plugging in $x$ directly, yields $0/0$ which is indeterminate. Applying L'Hopitals rule does not simplify the fraction, in fact the result is even more complicated and looks like a dead end, i.e. $\displaystyle{ \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}} = \lim_{x \to 0^-}{\frac{e^{1/x}}{x^2}}}$.
So I tried taking the natural log, i.e.
$\displaystyle{ y = \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}} }$

$\displaystyle{ \ln(y) = \lim_{x \to 0^-}{ \ln \left( \frac{e^{1/x}}{-x} \right) }
= \lim_{x \to 0^-}{ [1/x - \ln(-x) ]} = \lim_{x \to 0^-}{ \frac{1-x\ln(-x)}{x} } }$

Applying L'Hopitals Rule here gives me
$\displaystyle{ \lim_{x \to 0^-}{ -1-\ln(-x) } = +\infty }$ which is wrong; it should be $-\infty$.

I think I am making a very simple mistake but I don't see it.

Can anyone help?