For $x$ close to $0$, $1-2x$ is positive. So
$$(1-2x)^{1/x} = e^{\ln(1-2x)/x}.$$
Since the exponential function is continuous,
$$\lim_{x\to 0} e^{\ln(1-2x)/x} = e^{\scriptstyle\left(\lim\limits_{x\to 0}\ln(1-2x)/x\right)}$$
provided the latter limit exists. So this lets you change the original problem into the problem of determining whether
$$\lim_{x\to 0}\frac{\ln(1-2x)}{x}$$
exists, and if so what the limit is.
(Alternatively, since $\ln$ is continuous,
$$\lim_{x\to 0}\ln\left((1-2x)^{1/x}\right) = \ln\left(\lim_{x\to 0}(1-2x)^{1/x}\right)$$
so you can do the limit of the natural log instead).
Now, the limit
$$\lim_{x\to 0}\frac{\ln(1-2x)}{x}$$
is an indeterminate of type $\frac{0}{0}$, so you can try using L'Hopital's rule. We get
$$\begin{align*}
\lim_{x\to 0}\frac{\ln(1-2x)}{x} &= \lim_{x\to 0}\frac{(\ln(1-2x))'}{x'} &\text{(L'Hopital's Rule)}\\
&= \lim_{x\to 0}\frac{\quad\frac{1}{1-2x}(1-2x)'\quad}{1}\\
&= \lim_{x\to 0}\frac{(1-2x)'}{1-2x} \\
&= \lim_{x\to 0}\frac{-2}{1-2x}\\
&= -2.
\end{align*}$$
Hence
$$\begin{align*}
\lim_{x\to 0}(1-2x)^{1/x} &= \lim_{x\to 0} e^{\ln(1-2x)/x}\\
&= e^{\lim\limits_{x\to 0}\ln(1-2x)/x}\\
&= e^{-2}.
\end{align*}$$