First I found the Domains and Ranges of $f(x)$ and $g(x)$.
$D_{(f)} = \mathbb{R} -\{-1\}$
$R_{(f)} = \mathbb{R} -\{-1\}$
$D_{(g)} = \mathbb{R}$
$R_{(g)} = (-\infty, 1]$
Then I found $(g\circ f)_{(x)}$.
$(g\circ f): \mathbb{R}-\{-1\} \to (-\infty, 1]$
$(g\circ f)_{(x)} = \frac{8x(1-x)}{(1+x)^2}$
The problem is in finding $(f\circ g)_{(x)}$. Since $R_{g} = (-\infty, 1]$ and $D_{f} = \mathbb{R} -\{-1\}$, $\:g_{(x)}$ may return $-1$. To prevent that should I do as below ?
$(f\circ g)_{(x)}: \mathbb{R}-\{r_1,r_2\}\to \mathbb{R}-\{-1\}$; where $r_1, r_2$ are roots of the equation $g(x) = -1$
Since $\ D_\left(f\right)$ = $\ \mathbb{R} - \{1\} $ so we must exclude the point $\ -1$ from $\ R_{(g)}=(-\infty, 1]$.
Now $ g(x) = -1 \implies 4x(1-x)=-1\implies x= \frac{1\pm\sqrt{2}}{2}$
So $\ D_{(f\circ g)} = \mathbb{R}- \{ { \frac{1\pm\sqrt{2}}{2} }\} $.
So your perception is correct.