Show that for $\;\dfrac12I have the problem of unique. The existence is follows since using sided LHospital
$$\begin{cases}\lim\limits_{x\to0^+}f(x)=1\\{}\\\lim\limits_{x\to1^-}f(x)=\dfrac12\end{cases}\implies \;\text{the IVT for continuous functions proves}$$ .
For unique I did: evaluate differential
$$f'(x)=-\frac1{x\log^2x}+\frac1{(x-1)^2}=\frac{x\log^2x-(x-1)^2}{x(x-1)^2\log^2x}$$
Now, if I show $\;f'<0\;$ then $\;f\;$ is monotone going down and there is unique solution, but I'm having lots of problem for this: the denominator is always positive, but I can't prove numerator is negative.
After lots of tryings I tried direct proof:
$$f(x)=f(y)\iff \frac{\log\dfrac yx}{\log x\log y}=\frac{x-y}{(x-1)(y-1)}$$
and I won't have way to show $\;x=y\;$ .
Any helps is greatly appreciated.

In order to prove that
$$ f(x)=\frac{1}{\log x}-\frac{1}{x-1}$$
is decreasing over $(0,1)$, we just need to prove that:
$$ g(t) = \frac{1}{t}-\frac{1}{e^t-1} $$
is decreasing over $\mathbb{R}^-$. It is sufficient to prove that:
$$ \forall t<0,\quad g'(t) = \frac{e^t}{(1-e^t)^2}-\frac{1}{t^2}<0,$$
or:
$$ \forall t<0,\quad e^t+e^{-t}>2+t^2 $$
that is equivalent to:
$$ \forall t<0,\quad \cosh t > 1+\frac{t^2}{2} $$
that is trivial, since for any $t$:
$$ \cosh t = \sum_{n\geq 0}\frac{t^{2n}}{(2n)!} = 1+\frac{t^2}{2}+\frac{t^4}{24}+\ldots.$$
