Is there any way to represent the derivative of this complex quadratic statement into a compact matrix form?
$${x^{{*^T}}}Ax = \left[ {\begin{array}{*{20}{c}} {{a_1}{e^{ - j{\theta _1}}}}& \cdots &{{a_n}{e^{ - j{\theta _n}}}} \end{array}} \right]A\left[ {\begin{array}{*{20}{c}} {{a_1}{e^{j{\theta _1}}}} \\ \vdots \\ {{a_n}{e^{j{\theta _n}}}} \end{array}} \right]$$
For starters, the (Wirtinger) derivatives of $x^H A x$ (using $(\cdot)^H$ for conjugate transpose) with respect to $x$ and $x^*$ are just $A^T x^*$ and $A x$, respectively. If you want to derive with respect to $\theta$, use the chain rule:
$$\frac{\partial x^H A x}{\partial \theta} = \frac{\partial x^*}{\partial \theta}\cdot\frac{\partial x^H A x}{\partial x^*}+\frac{\partial x}{\partial \theta}\cdot\frac{\partial x^H A x}{\partial x} = \frac{\partial x^*}{\partial \theta} A x + \frac{\partial x}{\partial \theta} A^T x^* .$$
In your case, $\frac{\partial x^*}{\partial \theta}$ is simple, it is just a diagonal matrix with $-j a_i {\rm e}^{-j \theta_i}$ on its diagonal ($i=1, 2, ..., n$). For the derivative with respect to $a$, the same principle applies. Your partial derivatives are nothing but ${\rm e}^{-j \theta_i}$ in this case.