Plotting the function $\frac{x-1}{1+e^{\frac{1}{x-1}}}$

Question

I tried plotting the function $\frac{x-1}{1+e^{\frac{1}{x-1}}}$. For this I tried to gather information as follows:

1. The curve doesn't pass through origin.
2. When $x \rightarrow +\infty$, $y \rightarrow +\infty$
3. When $x \rightarrow -\infty$, $y \rightarrow -\infty$
4. When $x=0, y=-0.73$ and when $y=0, x=1$.

With this information, I drew this graph

To check if it is right, I plotted on desmos and it seems different. So I decided to go further.

5. To find the point of maxima/minima, inflection, I tried calculating the first and second derivative, \begin{equation*} \begin{split} y' &= \frac{1+e^{\frac{1}{x-1}}+e^{\frac{1}{x-1}}\times\frac{1}{x-1}}{(1+e^{\frac{1}{x-1}})^2} \end{split} \end{equation*}

6. To find maxima or minima, I have to equate $y'=0$

\begin{equation*} \begin{split} 1+e^{\frac{1}{x-1}}+e^{\frac{1}{x-1}}\times\frac{1}{x-1} &= 0\\ x-1 + x e^{\frac{1}{x-1}} &= 0 \end{split} \end{equation*}

How to solve this equation for value of $x$? From the graph on desmos, there is indeed a tangent with slope $0$ at the interval $[1,1.3]$. And believe it should be solvable.

7. To find the concavity direction, I know I have to evaluate $y''$. But it gets messier to have $y''$. From the graph on desmos again, the curve is concave upward both on the left and right of $x=1$.

Any help on how to solve points $6$ and $7$?