+118723 Mmm I am not great at these either but here is my logic.
\(f(x)=x^2-1\;\;for\;\;x<0\\ g(x)=\frac{1}{2x+1}\;\;for \;\;x<-0.5\\~\\ f(g(x))=(\frac{1}{2x+1})^2+1\qquad \text{domain to be determined}\\~ \\for\\ f(g(x)) \quad \text{Firstly } x <-0.5\\ but \;\;also\\ \frac{1}{2x+1}<0\\ \text{Yes that must already be true, no more restrictions necessary}\\ \text{So the domain will be }x<-0.5\\ \text{Now I will look at the range.}\\ \)
As x tends to -0.5 from below f(g(x)) tends to infinity
As x tends to -infinity f(g(x)) tends to -1
So the range is (-1, +infinity)
I think that is the same as you got.
And yes your last statement is correct.
But it was necessary to think about whether any further restraints were necessary for the inverse to still be a function. (no restraints are necessary)
Here are the graphs to back up our findings.
https://www.desmos.com/calculator/o0avvsjz6h
Some of your logic seemed simpler than mine but I am not sure if it was entirely trustworthy.
LaTex
f(x)=x^2-1\;\;for\;\;x<0\\
g(x)=\frac{1}{2x+1}\;\;for \;\;x<-0.5\\~\\
f(g(x))=(\frac{1}{2x+1})^2+1\qquad \text{domain to be determined}\\~
\\for\\
f(g(x)) \quad \text{Firstly } x <-0.5\\
but \;\;also\\
\frac{1}{2x+1}<0\\
\text{Yes that must already be true, no more restrictions necessary}\\
\text{So the domain will be }x<-0.5\\
\text{Now I will look at the range.}\\
!
