= Solution
Choose an analytic square-root branch of $f$ near infinity in a sector. The leading <Liouville–Green approximation> is
$$
w_\pm(z)\sim f(z)^{-1/4}\exp\left(\pm\int^z\sqrt{f(s)}ds\right),
$$
with successive inverse-power corrections; the two signs give an asymptotic basis in suitable sectors. Since $\sqrt f=\sqrt{f_0}+f_1/(2\sqrt{f_0}z)+\cdots$, the exponent contains a linear term and a logarithm. For the specified $f$, $\sqrt f=1+1/(2z)+7/(8z^2)+\cdots$, explaining $e^zz^{1/2}$ and $e^{-z}z^{-1/2}$ on the positive real axis.
To determine the requested coefficient from the differential equation itself, put $w_1=e^zz^{1/2}h(z)$. Substitution gives
$$
h''+(2+1/z)h'-\frac9{4z^2}h=0.
$$
For $h=1+a_1/z+\cdots$, the order-$z^{-2}$ coefficient is $-2a_1-9/4$. It must vanish, giving
$$
\boxed{a_1=-9/8}.
$$
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