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

Rewrite the equation $$x\tan\!\left(\frac{y}{x}\right)dy = \left(y\tan\!\left(\frac{y}{x}\right)-x\right)dx$$ as: $$\tan\!\left(\frac{y}{x}\right)(x\,dy - y\,dx) = -x\,dx.$$

Dividing both sides by $$x^2$$: $$\tan\!\left(\frac{y}{x}\right)d\!\left(\frac{y}{x}\right) = -\frac{dx}{x}.$$

Integrating both sides: $$-\ln\!\left|\cos\frac{y}{x}\right| = -\ln|x| + C$$, so $$\cos\frac{y}{x} = Kx$$ for some constant $$K$$.

Applying $$y\!\left(\frac{1}{2}\right) = \frac{\pi}{6}$$: $$\cos\frac{\pi/6}{1/2} = \cos\frac{\pi}{3} = \frac{1}{2} = K \cdot \frac{1}{2}$$, so $$K = 1$$. The solution is $$\cos\!\left(\frac{y}{x}\right) = x$$, i.e., $$y = x\arccos(x)$$.

The required area is $$\int_0^{1/\sqrt{2}} x\arccos(x)\,dx$$. Using integration by parts with $$u = \arccos(x)$$, $$dv = x\,dx$$: $$= \left[\frac{x^2}{2}\arccos x\right]_0^{1/\sqrt{2}} + \int_0^{1/\sqrt{2}} \frac{x^2}{2\sqrt{1-x^2}}\,dx.$$

The boundary term: $$\frac{1}{4}\cdot\frac{\pi}{4} = \frac{\pi}{16}$$.

For the remaining integral, substitute $$x = \sin\theta$$: $$\int_0^{\pi/4}\frac{\sin^2\theta}{2}\,d\theta = \frac{1}{4}\int_0^{\pi/4}(1-\cos 2\theta)\,d\theta = \frac{1}{4}\!\left[\theta - \frac{\sin 2\theta}{2}\right]_0^{\pi/4} = \frac{1}{4}\!\left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{16} - \frac{1}{8}.$$

Therefore the area is $$\frac{\pi}{16} + \frac{\pi}{16} - \frac{1}{8} = \frac{\pi}{8} - \frac{1}{8} = \frac{1}{8}(\pi-1)$$.