How to determine regions of positive and negative concavity of $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$? We may assume that $a<0,b<0, |b|>|a|$
How to check where the concavity of the function $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$ is positive or negative?
$h''(g)=\frac{4 \, {\left(a^{3} b + 2 \, a^{2} b^{2} + a b^{3} - {\left(a + b\right)} g^{3} + 3 \, {\left(a^{2} b + a b^{2}\right)} g\right)}}{a^{3} b^{3} + 3 \, {\left(a + b\right)} g^{5} + g^{6} + 3 \, {\left(a^{2} + 3 \, a b + b^{2}\right)} g^{4} + {\left(a^{3} + 9 \, a^{2} b + 9 \, a b^{2} + b^{3}\right)} g^{3} + 3 \, {\left(a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right)} g^{2} + 3 \, {\left(a^{3} b^{2} + a^{2} b^{3}\right)} g}$
The second derivative seems unhelpful.
Asymptote at $g=-a$, and $g=-b$

I don't think there is anything simpler (at least not to come up with) than just checking the second derivative. Here is the second derivative fully factored (over $\mathbb{Q}$):
$$\frac{4 (a+b) \left(a^2 b+a b^2+3 a b g-g^3\right)}{(a+g)^3(b+g)^3}$$
The main issue is that $a^2 b+a b^2+3 a b g-g^3$ is irreducible (except possibly for special values of $a, b$). Thus solving $h''(g) \ge 0$ becomes complicated. But we know that $h$ is convex iff $h'' \ge 0$ so those roots are going to show up no matter what method we try to use.
The discriminant of that cubic is $-27 a^2 (a - b)^2 b^2$. So there is a single real root (multiplicity $1$) so long as $a, b, a - b \neq 0$.
