In how many points is tangent of $f(x)=\frac{e^x-2}{(e^x-1)^2}$ parallel with $y=x$ In how many points is tangent of $f(x)=\frac{e^x-2}{(e^x-1)^2}$ parallel with $y=x$

I found first derivative and put $$f'(x)=\frac{-e^x(e^x-3)}{(e^x-1)^3}=1$$

I can't solve it. I think that trick is only to find how many solutions are there, not finding exact ones.
Hint:

Let $y=e^x$.

Then the equation symplifies to a polynomial equation of $-y(y-3)=(y-1)^3$.

Hint2:

You don't need to solve this equation, you only need to show how many solutions it has.
One possibility to show how many solutions an equation has, is to use the intermediate value theorem.

https://en.wikipedia.org/wiki/Intermediate_value_theorem

you know your function is continous and look for points where your new function $g(x):=f(x)-x$ is $&gt;0$ and $&lt;0$; intermediate value theorem says then there must be a solution.