Solution:

Answer: $-\frac{3}{8}$

Viewing $x$ as a constant and completing the square, we find that
$$\begin{aligned} z & = 4x^2 - 4x y + y^2 - 2y^2 - 3y \\ & = -y^2 - (4x + 3) y + 4x^2 \\ & = -\left(y + \frac{4x + 3}{2}\right)^2 + \left(\frac{4x + 3}{2}\right)^2 + 4x^2 \end{aligned}$$
Brahmagupta wishes to maximize $z$, so regardless of the value of $x$, he will pick $y = -\frac{4x + 3}{2}$. The expression for $z$ then simplifies to
$$z = 8x^2 + 6x + \frac{9}{4}$$
Archimedes knows this and will therefore pick $x$ to minimize the above expression. By completing the square, we find that $x = -\frac{3}{8}$ minimizes $z$.

Alternatively, note that $z$ is convex in $x$ and concave in $y$, so we can use the minimax theorem to switch the order of moves. If Archimedes goes second, he will set $x = \frac{y}{2}$ to minimize $z$, so Brahmagupta will maximize $-2y^2 - 3y$ by setting $y = -\frac{3}{4}$. Thus Archimedes should pick $x = -\frac{3}{8}$, as above.