Well, I also couldn't understand Thierry's argument, so I decided to use the old "pushing junk down the ladder" trick from classical asymptotic analysis. Note first of all that looking at the expression for the *first* derivative, which is an integral of a non-negative quantity, we can easily realize that either $f'(t)$ is identically $0$, or $f'(t)\ge ce^{-Ct}$ (I prefer to use $t$ instead of $\gamma$ for typing reasons). Canceling $e^{-{y^2}/{2}}$ whenever possible and changing the variable $y$ to $\sqrt ty$, we see that up to some irrelevant factor, $f''(t)$ is the same as
$$
F(t)=\int_{-\infty}^\infty e^{-t\frac{y^2}2}\frac{\sum_k\alpha_k e^{t U_k(y)}}{\sum_j\beta_j e^{t V_j(y)}}\,dy
$$
where $U_k,V_j$ are some fixed linear functions of $y$ and the sums are finite. Note that $I(t)\le e^{Ct}$ for some $C>0$ (in your case, it has no chance to grow at all, but let's be generous and use the particular structure of your function only when we need it). Now look at $W(y)=\max_j V_j(y)$. It is a piecewise linear function. Note that the whole real line can be split into finitely many (bounded or unbounded) intervals  $J_m$, each interval containing one point where the graph of $W$ has a corner so that on each interval we have a few linear forms $V_j$ meeting at one point while all the rest stay down from the maximum of these meeting forms by some constant. Splitting at that corner point and moving the orijin to it, we see that $F(t)$ is a sum of integrals of the form
$$
I(t)=\int_0^a e^{-t\frac{y^2}2}\frac{S(t,y)}{R(t,y)+J(t,y)}\,dy
$$
where $S(t,y)$ and $J(t,y)$ are of the same form $\sum_k\alpha_k e^{t U_k(y)}$ with arbitrary linear forms $U_k$, while $R(t,y)=1+\sum_j{\beta_je^{-\lambda_j t}}$ with some positive $\lambda_j$ and $J(t,y)\le e^{-ct}R(t,y)$ on $[0,a)$.

Now it is time to push the junk term $J(t,y)$ down the ladder by writing
$$
\frac 1{R(t,y)+J(t,y)}=\frac 1{R(t,y)}\sum_{m=0}^M (-1)^m\left[\frac{J(t,y)}{R(t,y)}\right]^m+\frac 1{R(t,y)+J(t,y)}(-1)^{M+1}\left[\frac{J(t,y)}{R(t,y)}\right]^{M+1}
$$ 
If $M$ is chosen large enough, the last term we want to drop is dominated by $e^{-Ct}$ times the original expression, so, stopping far enough, we can make the error decay exponentially with as large exponent as we wish. Thus, we can get an approximation, which is a linear combination of terms of the kind
$$
T(t)=e^{bt}\int_0^a e^{-t\frac{y^2}2}\frac{e^{\lambda ty}}{1+Q(t,y)}\,dy
$$
where $Q(t,y)=\sum_j{\beta_je^{-\lambda_j t}}$ with some positive $\lambda_j$. 

Now we have to consider 2 cases.

Case 1: There is no $Q(t,y)$ (pure exponent).
Here we can shift everything to the point of maximum of $\lambda y-\frac{y^2}2$, split if necessary, and get a linear combination of terms of the kind 
$$
e^{bt}\int_0^a e^{-t\frac{y^2}2}\,dy=ce^{bt}t^{-1/2}-ce^{b't}\int_0^{\infty}e^{-t\frac{y^2}2}e^{-aty}\,dy
$$
(write $\int_0^a=\int_{0}^\infty-\int_{a}^\infty$ and shift the starting point back to the origin in the second one).

Case 2: There is a non-trivial $Q(t,y)$ in the game. Then we want to push the junk down the ladder once again writing 
$$
\frac{1}{1+Q(t,y)}=\sum_{m=0}^M (-Q(t,y))^m+\frac{(-Q(t,y))^{M+1}}{1+Q(t,y)}
$$
The purpose of the game is to get a lot of pure exponents and an *exponentially decaying* function $e^{\lambda t}Q(t,y)^{M+1}$ in the numerator (note that, for all we know, $\lambda$ could be large positive in the beginning). Moreover, when integrating the last term, we can add the integral from $a$ to $\infty$ because that one can be made to decay exponentially with arbitrarily large exponent by choosing $M$ large enough.

Now, with all the junk pushed ruthlessly away, we get the sum of the terms of two kinds:
$e^{bt}t^{-1/2}$ and $e^{bt}\int_0^\infty e^{-t\frac{y^2}2}f(ty)\,dy$ where $f$ is some exponentially decaying function on $(0,\infty)$. Plus, of course, we have a very quickly decaying error. Of course, we can combine the terms of each kind with the same $b$, so I'll assume that all $b$'s are different.

In the terms of the second kind, one is tempted to change variable to $ty$ getting
$$
t^{-1}\int_0^\infty e^{-t^{-1}\frac{y^2}2}f(y)\,dy
$$
At this point, we can safely write the asymptotic expansion in the inverse powers of $t$ just expanding $e^-z$ and integrating formally. Note that it will be merely an asymptotic expansion, not a convergent series. However, it is enough for us to get the asymptotics for the whole thing unless all coefficient miraculously vanish. However, the coefficients are even moments of an exponentially decaying function. Reflecting it around the origin, we get a function with all vanishing moments and exponential decay. The Fourier transform of such function will be a real analytic function with all derivatives at the origin equal to $0$. Thus, it is identically $0$, and so is the original $f$. But then we have total collapse of all terms and only the error is present, i.e., $F$ decays faster than any exponent. It is, probably, impossible in general but in your particular case, we have a clear reason: if $f''$ decays superexponentially, so does $f'$, but we ruled that out from the beginning.

Of course, if we can fix the Pfaffian argument, this "junk" method will become totally obsolete. But until Thierry replies, one can live with it :).