1. Write the curves in convenient form

The given curves are:

$x(1+y^2)=1 \quad \Rightarrow \quad x=\frac{1}{1+y^2}$

and

$y^2=2x \quad \Rightarrow \quad x=\frac{y^2}{2}.$

So for a given $y$, the region is bounded horizontally between

$x=\frac{y^2}{2} \quad \text{and} \quad x=\frac{1}{1+y^2}.$

2. Find the points of intersection

At intersection,

$\frac{1}{1+y^2}=\frac{y^2}{2}.$

Multiply through:

$2=y^2(1+y^2)$

$y^4+y^2-2=0.$

Let $t=y^2$. Then

$t^2+t-2=0=(t-1)(t+2).$

Since $t=y^2\ge 0$, we get

$t=1 \Rightarrow y^2=1 \Rightarrow y=\pm 1.$

Corresponding $x$-value:

$x=\frac{y^2}{2}=\frac12.$

Thus the curves intersect at

$(\tfrac12,1), \quad (\tfrac12,-1).$

3. Set up the area integral

For $-1\le y\le 1$,

$\frac{1}{1+y^2} \ge \frac{y^2}{2},$

so area is

$A=\int_{-1}^{1}\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy.$

Since the integrand is even,

$A=2\int_0^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy.$

4. Evaluate the integral

$A=2\left[\int_0^1 \frac{1}{1+y^2}\,dy-\int_0^1 \frac{y^2}{2}\,dy\right].$

Now,

$\int_0^1 \frac{1}{1+y^2}\,dy=\left[\tan^{-1}y\right]_0^1=\frac{\pi}{4},$

and

$\int_0^1 \frac{y^2}{2}\,dy=\frac12\left[\frac{y^3}{3}\right]_0^1=\frac16.$

Therefore,

$A=2\left(\frac{\pi}{4}-\frac16\right)=\frac{\pi}{2}-\frac13.$

5. Match with the options

$\boxed{A=\frac{\pi}{2}-\frac13}$

This corresponds to Option B.