to find how is the behavior of h(x) we have to calculate it's derivative so:
$$h' = \frac{f'g-g'f}{g^2}$$
in order that this fraction be positive(so h be increasing) we must have:
$$f'g-g'f\ge0 \;\;\;\;\; \; \Rightarrow \frac{f'}{f} \ge \frac{g'}{g}$$
now integrating yields:
$$ln(f)\ge ln(g) + constant \Rightarrow f \ge constant \times g $$
so you must identify that f is always greater than g and since g and f are both positive this test can be done easily. on the other hand if
$$g \ge constant \times f $$
then h is decreasing.
this just popped into my head now but there must be more sophisticated ways.
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After discussion with S.harp he proved to me the above statement is wrong and the counter example is the one in the comments below. so I edit this answer this way:
that constant is an arbitrary positive number. so for above solution to be true there must be an interval of the form $(\beta,+ \infty) \; $ in which the inequality $$f \ge constant \times g $$ or the other one holds for any positive arbitrary constant . this happens when the growth rate of f is greater than g other whise the claim could be wrong. for example: $f= 100+x \; and \; g = x$ although $f>g$ but $h = f/g $ is decreasing. so if growth rate of f is greater than g then the claim is right.
I apologize for wrong response.
also thanks to S.harp for revealing that first claim was wrong.
One keeps all definitions as in the question.
Lemma. Let $I$ be an interval either of the form $(\alpha ,0]$ or $[0,\beta)$. Let $J:=I\setminus \{0\}$. The $f'(J)$ contains the interval $(-1,1)$.
Proof. It is clear that $$\limsup_{x\rightarrow 0^+}f'(x)=1,\liminf_{x\rightarrow 0^+}f'(x)=-1,$$ and similarly
$$\limsup_{x\rightarrow 0^-}f'(x)=1,\liminf_{x\rightarrow 0^-}f'(x)=-1.$$
One prove the case when $J=(0,\beta)$ (the other case being similar). If $y_0\in (-1,1)$, then $y_0\in (-1+\epsilon,1-\epsilon)$ for som positive $\epsilon$. From the above observations, there exists $x_2\in J$ such that $f'(x_2)>1-\epsilon$. And then there exists $0<x_1<x_2$ such that $f'(x_1)<-1+\epsilon$. Now since $(x_1,x_2)\subseteq J$ and $f'$ is continuous there, there exists by IVP for $f'$ that $\exists x_0\in (x_1,x_2)$ with $f'(x_0)=y_0.$ This completes the proof.
Proposition. $f'$ satisfies the IVP, namely for $I=[a,b]$, $f'$ assumes any value $y$ between $f'(a)$ and $f'(b)$.
Proof. As the OP remarked, it suffices to prove the case when $0\in I$ (so $a\leq 0,b\geq 0$). Also the case when $f'(a)=f'(b)$ being void, one assumes that $f'(a)\neq f'(b).$ Let $y$ be any value between $f'(a)$ and $f'(b)$. Now one may divide into two cases.
Case 1. $|y|<1.$
This follows from the lemma above.
Case 2. There are $4$ subcases (possibly overlapping) as follows (where one applies the lemma above to choose $x_0$):
Subcase 1. $1\leq y<f'(b)$: Take $0<x_0<b$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[x_0,b]$.
Subcase 2. $1\leq y<f'(a)$: Take $a<x_0<0$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 3. $f'(a) < y\leq -1$: Take $a<x_0<0$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 4. $f'(b)<y\leq -1$: Take $0<x_0<b$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[x_0,b]$.
Combining all cases, the proposition is proven.
Best Answer
I will assume the properties $e^x > 1 +x + \frac{x^2}{2}$ and $e^{x+y} = e^x e^y$ are given.
For $x+h > x > 0$, we have $\left(h+ \frac{h^2}{2}\right)\left(1+x+ \frac{x^2}{2}\right)> h + xh+\frac{h^2}{2}$, and, hence,
$$e^h > 1+ h + \frac{h^2}{2} > 1 + \frac{h + xh+\frac{h^2}{2}}{1+x+ \frac{x^2}{2}} = \frac{1 + (x+h)+\frac{(x+h)^2}{2}}{1+x+ \frac{x^2}{2}}$$
Thus,
$$1+x+ \frac{x^2}{2} > e^{-h}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Multiplying both sides by $e^{-x}$ we get
$$e^{-x}\left(1+x+ \frac{x^2}{2}\right) > e^{-(x+h)}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Something similar should work for the case $x < 0$.