= Solution
For fixed $\gamma$, <differentiation under the integral sign> is justified by bounded $P$ and $X$. At the unconstrained <minimizer> $H_\gamma$, the first-order condition is
$$
p\,e^{\gamma(H_\gamma\cdot p-x)}
=\mathbb E\!\left[P e^{\gamma(X-H_\gamma\cdot P)}\right].
$$
Define the strictly positive, bounded <pricing kernel>
$$
Z_\gamma=\frac{e^{\gamma(X-H_\gamma\cdot P)}}{e^{\gamma(H_\gamma\cdot p-x)}}.
$$
The first-order condition says $\mathbb E[Z_\gamma P]=p$, so $Z_\gamma\in\mathcal Z$. When taking the requested <partial derivative>, hold $H$ fixed and only afterwards set $H=H_\gamma$. Writing $A_\gamma=H_\gamma\cdot p-x$, we obtain
$$
\begin{aligned}
\left.\partial_\gamma F_\gamma(H)\right|_{H=H_\gamma}
&=e^{\gamma A_\gamma}\left[A_\gamma+
\mathbb E\{Z_\gamma(X-H_\gamma\cdot P)\}\right]\\
&=e^{\gamma A_\gamma}\big[\mathbb E(Z_\gamma X)-x\big].
\end{aligned}
$$
The assumed dual bound applies to this particular <pricing kernel>. \b[Thus]
$$
\boxed{\left.\partial_\gamma F_\gamma(H)\right|_{H=H_\gamma}\leq0.}
$$
No derivative of the map $\gamma\mapsto H_\gamma$ is needed; confusing a <partial derivative> with a derivative along the minimizing path would add an unnecessary hypothesis.
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