Some "Log"ic required

$\begin{aligned} \log_4 (x+2y) + \log_4(x-2y) & = 1 & \small \color{#3D99F6} \text{Note that }\log AB = \log A + \log B \\ \log_4 (x+2y)(x-2y) & = 1 \\ \log_4 (x^2-4y^2) & = 1 & \small \color{#3D99F6} \text{and }\log_b a = c \implies a = c^b \\ x^2-4y^2 & = 4 \\ \implies x & = \pm 2 \sqrt{y^2 + 1} \end{aligned}$

Therefore, we have:

$\begin{aligned} z & = |x| - |y| = 2\sqrt{y^2+1} - y & \small \color{#3D99F6} \text{For }y > 0 \\ \frac {dz}{dy} & = \frac {2y}{\sqrt{y^2+1}} - 1 = \frac {2y-\sqrt{y^2+1}}{\sqrt{y^2+1}} \end{aligned}$

Putting $\dfrac {dz}{dy} = 0$ $\implies 2y = \sqrt{y^2+1}$ $\implies 4y^2 = y^2 + 1$ $\implies y = \pm \dfrac 1{\sqrt 3}$ . Note that $\dfrac {d^2z}{dy^2} = \dfrac {2-2y^2} {(y^2+1)^\frac 32}$ and $\dfrac {d^2z}{dy^2} > 0$ when $y = \pm \dfrac 1{\sqrt 3}$ . This implies that $\min (z) = 2\sqrt{\frac 13+1} - \frac 1{\sqrt 3} = \sqrt 3 \approx \boxed{1.732}$ .

Hint :- Parameterise the hyperbola..........!!!