Let \( y=y(x) \) be the solution of the differential equation \( x\tan\!\left(\dfrac{y}{x}\right)dy=\left(y\tan\!\left(\dfrac{y}{x}\right)-x\right)dx, \; -1\le x\le 1, \; y\!\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6} \). Then the area of the region bounded by the curves \( x=0 \), \( x=\dfrac{1}{\sqrt{2}} \) and \( y=y(x) \) in the upper half plane is:
