Let $\vec\nabla f$ be a $n \times 1$ vector, we will call this $\dfrac{\partial f}{\partial x}$. Then we get that $\dfrac{\partial f}{\partial x} = Ax + A^Tx + b.$ The differentiation with respect to a vector is obtained by recalling the following vector differentiation identities.
If $y \in \mathbb{R}^{m \times 1}$, $x \in \mathbb{R}^{n \times 1}$, then $\dfrac{\partial y^T}{\partial x}$ is a $n \times m$ matrix $A$ with $A(i,j) = \dfrac{\partial y_j}{\partial x_i}$.
Also, the chain rule goes as follows. $\dfrac{\partial (\cdot)}{\partial x} = \sum_k \dfrac{\partial y_k^T}{\partial x}\dfrac{\partial (\cdot)}{\partial y_k}$ Hence, we have the following relations $\dfrac{\partial x^T}{\partial x }= I_{n \times n}$ $\dfrac{\partial (c^Tx)}{\partial x} = \dfrac{\partial (x^Tc)}{\partial x} = c$
Hence, $\dfrac{\partial b^Tx}{\partial x} = b$ and \begin{align} \dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial x} & = \dfrac{\partial y_1^T}{\partial x}\dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x}\dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_2}\\ & = \dfrac{\partial y_1^T}{\partial x} \dfrac{\partial \left(y_2(x)^TA^Ty_1(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x} \dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_2}\\ & = \dfrac{\partial y_1^T}{\partial x} \dfrac{\partial \left((Ay_2(x))^Ty_1(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x} \dfrac{\partial \left((A^Ty_1(x))^Ty_2(x) \right)}{\partial y_2}\\ & = \dfrac{\partial y_1^T}{\partial x} Ay_2(x) + \dfrac{\partial y_2^T}{\partial x} A^Ty_1(x) \end{align} In our case, $y_1(x) = y_2(x) = x$ and hence $\dfrac{\partial y_1^T}{\partial x} = \dfrac{\partial y_2^T}{\partial x} = I_{n \times n}$. Hence, $\dfrac{\partial (x^TAx)}{\partial x} = I A x + I A^T x = \left(A+A^T \right)x$ Setting the gradient to zero, we get the equation $(A+A^T)x = -b$ Solve this linear system to get the stationary points.