first you can find the critical points of $f$ (under which the inner extrema occur) in $(0, A) \times (A, \infty)$ by solving $ \frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial y}(x,y) = 0. $ After that find the critical points of $f(\cdot, A)$, $f(0, \cdot)$ and $f(A, \cdot)$ by solving $ \frac{\partial f}{\partial x}(x, A) = 0, \frac{\partial f}{\partial y}(0, y) = 0 \text{ resp. } \frac{\partial f}{\partial y}(A, y) $ Then compute the values of $f$ at each critical point and $f(0,A)$ and $f(A, A)$. You have found your minimum.
Concernig your example: We have \begin{align*} \partial_x f(x,y) &= -2x\\ \partial_y f(x,y) &= 4y + 3 \end{align*} Hence there are no inner critical points, no critical points on $(0, A) \times \{A\}$ and no on $\{0, A\} \times (A, \infty)$. So we compute $f(0,1) = 2 + 3 + 8 = 13$ and $f(1,1) = -1 + 2 + 3 + 8 = 12$. The mimimal positive value is therefore 12 (as $f$ is increasing for $y \to \infty$).