= Solution
Differentiating the <gradient> term by term gives the <Hessian matrix>
$$
\nabla^2l(\mathbf x)=
\sum_{i=1}^n y_i f''(\mathbf a_i^{\mathsf T}\mathbf x)\,
\mathbf a_i\mathbf a_i^{\mathsf T}.
$$
Take row $i$ of $A$ to be $\mathbf a_i^{\mathsf T}$. Then
$$
\boxed{\nabla^2l=A^{\mathsf T}\Delta A,\qquad
\Delta_{ii}=y_i f''(\mathbf a_i^{\mathsf T}\mathbf x),\quad
\Delta_{ij}=0\ (i\ne j).}
$$
Indeed, entry $(j,k)$ of this product is $\sum_i\Delta_{ii}a_{ij}a_{ik}$, exactly the entry of the displayed sum of <outer products>. Explicitly,
$$
\Delta_{ii}=y_i\bigl((\mathbf a_i^{\mathsf T}\mathbf x)^2-1\bigr)
e^{-(\mathbf a_i^{\mathsf T}\mathbf x)^2/2}.
$$
These coefficients may have either sign, so a <positive-definite> <Hessian matrix> cannot be assumed. This is an instance of the <Hessian of a sum of functions of linear forms>.
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