Are you good in geometry?

The equation $2^{|x|}|y| + 2^{|x|-1} = 1$ is an even on both $x$ and $y$ . There equation has $4$ curves symmetrical on x- and y-axes. The area bounded by the inequality $2^{|x|}|y| + 2^{|x|-1} \le 1$ and $|x| \le \frac{1}{2}$ or $-\frac{1}{2} \le x \le \frac{1}{2}$ is $4 \times$ that of under the curve $2^x y + 2^{x-1} = 1$ for $0 \le x \le \frac{1}{2}$ . We note that $2^x y + 2^{x-1} = 1$ , $\Rightarrow y = \dfrac{1}{2^x} - \frac{1}{2}$ and that $y \le \frac{1}{2}$ for all $x$ . There the area $A$ bounded by the inequality $2^{|x|}|y| + 2^{|x|-1} \le 1$ , $|x| \le \frac{1}{2}$ and $\color{#3D99F6}{|y| \le \frac{1}{2}}$ is same as that bounded by the inequality and $|x| \le \frac{1}{2}$ . Therefore, we have:

$\begin{aligned} A & = 4 \int_0^{\frac{1}{2}} y \space dx \\ & = 4 \int_0^{\frac{1}{2}} \left(2^{-x} -\frac{1}{2}\right) dx \\ & = 4 \left[-\frac{2^{-x}}{\ln 2} - \frac{x}{2}\right]_0^{\frac{1}{2}} \\ & = 4 \left( \frac{1-\frac{1}{\sqrt{2}}}{\ln 2} - \frac{1}{4} \right) \approx \boxed{0.6902} \end{aligned}$