$|x|+|y| \leq 1$
$2 y^{2} \geq|x|$
For point of intersection
$x+y=1 \Rightarrow x=1-y$
$y^{2}=\frac{x}{2} \Rightarrow 2 y^{2}=x$
$2 y^{2}=1-y \Rightarrow 2 y^{2}+y-1=0$
$(2 y-1)(y+1)=0$
$y =\frac{1}{2}$ or -1
Now Area of $\Delta OAB =\frac{1}{2} \times 1 \times 1=\frac{1}{2}$
Area of Region $R _{1}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$
Area of Region $R _{2}=\frac{1}{\sqrt{2}} \int\limits_{0}^{\frac{1}{2}} \sqrt{ x } dx =\frac{1}{6}$
Now area of shaded region in first quadrant
$=$ Area of $\Delta OAB - R _{1}- R _{2}$
$=\frac{1}{2}-\left(\frac{1}{6}\right)-\left(\frac{1}{8}\right)=\frac{5}{24}$
So required area $=4\left(\frac{5}{24}\right)=\frac{5}{6}$
