The area of the region bounded by the curves $$x(1+y^{2})=1$$ and $$y^{2}=2x$$ is:

We need to find the area of the region bounded by the curves $$x(1+y^2) = 1$$ and $$y^2 = 2x$$.

We start by finding the intersection points. From $$y^2 = 2x$$: $$x = \frac{y^2}{2}$$ and from $$x(1+y^2) = 1$$: $$x = \frac{1}{1+y^2}$$. Setting these equal, $$\frac{y^2}{2} = \frac{1}{1+y^2}$$, which gives $$y^2(1+y^2) = 2 \implies y^4 + y^2 - 2 = 0 \implies (y^2+2)(y^2-1) = 0$$. Since $$y^2 \geq 0$$, it follows that $$y^2 = 1$$ and hence $$y = \pm 1$$. At $$y = \pm 1$$, $$x = \frac{1}{2}$$.

Next, we determine which curve is to the right. For $$|y| < 1$$, $$\frac{1}{1+y^2} > \frac{y^2}{2}$$ (e.g., at $$y=0$$: $$1 > 0$$), so $$x = \frac{1}{1+y^2}$$ lies to the right of $$x = \frac{y^2}{2}$$.

Then we compute the area: $$\text{Area} = \int_{-1}^{1}\left[\frac{1}{1+y^2} - \frac{y^2}{2}\right]dy = 2\int_{0}^{1}\left[\frac{1}{1+y^2} - \frac{y^2}{2}\right]dy$$ (by symmetry about the x-axis), which equals $$2\left[\arctan y - \frac{y^3}{6}\right]_0^1 = 2\left[\frac{\pi}{4} - \frac{1}{6}\right] = \frac{\pi}{2} - \frac{1}{3}$$.

Rewriting, $$\frac{\pi}{2} - \frac{1}{3} = \frac{3\pi - 2}{6}$$.

Comparing with the options, Option A is $$2\left(\frac{\pi}{2} - \frac{1}{3}\right) = \pi - \frac{2}{3}$$, Option B is $$\frac{\pi}{2} - \frac{1}{3}$$, Option C is $$\frac{\pi}{4} - \frac{1}{3}$$, and Option D is $$\frac{1}{2}\left(\frac{\pi}{2} - \frac{1}{3}\right) = \frac{\pi}{4} - \frac{1}{6}$$. Option B matches.

The area of the region is $$\frac{\pi}{2} - \frac{1}{3}$$.