= Solution
Write the contour near the simple saddle as
$$
z=z_0+e^{i\alpha}s,\qquad s\in\mathbb R,
$$
with its given orientation. Since $\phi'(z_0)=0$,
$$
\phi(z)
=\phi(z_0)+\frac12\phi''(z_0)e^{2i\alpha}s^2
+O(s^3).
$$
Along a steepest-descent tangent the quadratic coefficient is real and negative, so
$$
\phi''(z_0)e^{2i\alpha}=-|\phi''(z_0)|,
\qquad
\arg\phi''(z_0)+2\alpha=\pi\pmod{2\pi}.
$$
Moreover $dz=e^{i\alpha}ds$. Only a neighbourhood of width $O(x^{-1/2})$ contributes at leading order, and there $f(z)=f(z_0)+o(1)$. The local integral is consequently
$$
f(z_0)e^{x\phi(z_0)+i\alpha}
\int_{-\infty}^{\infty}
\exp\left(-\frac{x|\phi''(z_0)|}{2}s^2\right)\,ds.
$$
Evaluating the Gaussian proves the <simple-saddle contribution in steepest descent>:
$$
\boxed{
I(x)\sim
f(z_0)\sqrt{\frac{2\pi}{x|\phi''(z_0)|}}\,
e^{x\phi(z_0)+i\alpha}}.
$$
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