We have the following definitions:
\begin{equation*}
V^*:=\{f\in S^*\colon\exists x\in S\ f(x)=1\},
\end{equation*}
where $S$ and $S^*$ are the unit spheres in the underlying Banach space $E$ and its dual $E^*$, respectively, and
\begin{equation*}
T(x):=\{f\in E^*\colon\|f\|\le1,f(x)=1\}
\end{equation*}
for $x\in S$.
We want to show that
\begin{equation*}
f\in V^*\overset{\text{(?)}}\implies\exists x\in S\ T(x)=\{f\} \tag{1}\label{1}
\end{equation*}
provided that the norm on $E$ is Gateaux differentiable.
Since $f\in V^*$, we have $\|f\|=1$ and $f(x)=1$ for some $x\in S$; fix any such $x$. Then
\begin{equation*}
f\in T(x).
\end{equation*}
Take any $g\in T(x)$. It suffices to show that $g=f$.
To obtain a contradiction, suppose that $g\ne f$. Then
\begin{equation*}
f(y)=0\ne g(y)
\end{equation*}
for some $y\in E$. Fix any such $y$.
By the Gateaux differentiability of the norm on $E$, for some real $l_y$ we have
\begin{equation*}
\|x+ty\|=\|x\|+l_y t+o(t)=1+l_yt+o(t)
\end{equation*}
(as $t\to0$).
Case 1: $l_y\ne0$. Then
\begin{equation*}
1=f(x)=f(x)+tf(y)=f(x+ty)\le\|x+ty\|=1+l_y t+o(t)<1
\end{equation*}
if $|t|$ is small enough and $l_y t<0$. So, Case 1 results in a contradiction. (Note that the condition $g(y)\ne0$ was not used in Case 1.)
Case 2: $l_y=0$. Then for any real $a\ne0$ and all real $t$ close enough to $0$ we have
\begin{equation*}
\|x+t(ax+y)\|=\|(1+ta)x+ty\|=(1+ta)\|x+\tfrac t{1+ta}\,y\| \\
=(1+ta)(1+0 t+o(t))=1+at+o(t),
\end{equation*}
so that
\begin{equation*}
l_{ax+y}=a\ne0.
\end{equation*}
Note also that $g(ax+y)=a+g(y)=0$ if $a=-g(y)\ne0$. So, replacing $y$ by $ax+y$ and interchanging the roles of $f$ and $g$, we reduce Case 2 to Case 1. $\quad\Box$