We have a generator $X$ that selects numbers from a uniform distribution on $(0,1)$ denoted $\text{Unif}(0,1)$.

We have to show how it can be used with the function $x \mapsto \log\left(\frac{x}{1-x}\right)$ to generate a random number from a distribution with cumulative distribution function $F(x)=\frac{e^x}{1+e^x}$.

The calculation required is to verify that $F^{-1}\left(X\right)$ has cumulative distribution function $F$.

I try it this way:

Let random variable $X \sim \text{Unif}(0,1)$.

$P\left(\log\left(\frac{X}{1-X}\right)\le x\right)=P\left(\frac{X}{1-X}\le e^x\right)=P\left(X\le \frac{e^x}{1+e^x}\right)$.

I think that this is the way but I don't know how to continue and how to justify that:

$P\left(X\le \frac{e^x}{1+e^x}\right)=\frac{e^x}{1+e^x}$.