$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\i{\mathbf{i}}$
Using the same parametrization as in the answer abut smoothness of the arithmetic mean,
we have
\begin{align}
\Wp(\alpha)&=-\tfrac\alpha2\,\cot\tfrac\alpha2-\tfrac\alpha2\cdot\i
\tag{1}
\end{align}
\begin{align}
\Wm(\alpha)
&=-\tfrac\alpha2\,\cot\tfrac\alpha2+\tfrac\alpha2\cdot\i
\tag{2}\label{2}
,
\end{align}
\begin{align}
x(\alpha)&=
-\frac{\alpha}{2\sin\tfrac\alpha2}
\cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2)
\tag{3}\label{3}
\end{align}
for $x\le-\tfrac1\e$,
$\alpha\ge0$, since
\begin{align}
\lim_{\alpha\to0}x(\alpha)=-\tfrac1\e
\tag{4}\label{4}
.
\end{align}
Combining this with
the parametric representation
of $\Wp(x),\,\Wm(x)$
we have
\begin{align}
\Wp(a)&=\frac{a\ln a}{1-a}
\tag{5}\label{5}
,\\
\Wm(a)&=\frac{\ln a}{1-a}
\tag{6}\label{6}
,\\
x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}}
\tag{7}\label{7}
\end{align}
for $x\in[-\tfrac1\e,0]$,
$a\in[0,1]$.
The geometric mean of interest, defined as
\begin{align}
f_g(x)&=-\sqrt{\Wp(x)\Wm(x)}
\tag{8}\label{8}
\end{align}
for $x<0$
is then represented as
two pieces
\begin{align}
f_{g1}(\alpha)&=-\frac{\tfrac\alpha2}{\sin\tfrac\alpha2}
\tag{9}\label{9}
,\\
x(\alpha)&=
-\frac{\alpha}{2\sin\tfrac\alpha2}
\cdot\exp(-\tfrac\alpha2\,\cot\tfrac\alpha2)
,\quad \alpha>0
\tag{10}\label{10}
,\\
\text{and }\quad
f_{g2}(a)&=\frac{\sqrt{a}\ln a}{1-a}
\tag{11}\label{11}
,\\
x(a)&=\frac{\ln a}{1-a}\cdot a^{\tfrac1{1-a}}
,\quad a\in[0,1]
\tag{12}\label{12}
,\\
\lim_{\alpha=0}x(\alpha)&=
\lim_{a=1}x(a)=
-\tfrac1\e
\tag{13}\label{13}
,\\
\lim_{\alpha=0}f_{g1}(\alpha)&=
\lim_{a=1}f_{g2}(a)=-1
\tag{14}\label{14}
.
\end{align}
Omitting details of calculation,
this leads to,
\begin{align}
\left.
\frac{d f_{g1}(\alpha)/d\alpha}{dx(\alpha)/d\alpha}
\right|_{\alpha=0}
&=
\left.
\frac{d f_{g2}(a)/da}{dx(a)/da}
\right|_{a=1}
=\frac13\e
\tag{15}\label{15}
,\\
\left.
\frac{d^2 f_{g1}(\alpha)/d\alpha^2}{d^2x(\alpha)/d\alpha^2}
\right|_{\alpha=0}
&=
\left.
\frac{d^2 f_{g2}(a)/da^2}{d^2x(a)/da^2}
\right|_{a=1}
=\frac13\e
\tag{16}\label{16}
,
\end{align}
that means that $f_g(x)$ is indeed $C^2$-continuous at $x=-\tfrac1\e$.
And again, exactly as in the case of the arithmetic mean,
the value of first and second derivation
at this point is the same.
$\endgroup$
