$$\begin{aligned} & x\left(1+y^2\right)=1 \quad\text{..... (1)}\\ & y^2=2 x\quad\text{..... (2)} \end{aligned}$$

$$\begin{aligned} &\text { From equation (1) & (2) }\\ &\begin{aligned} x(1+2 x)=1 & \Rightarrow 2 x^2+x-1=0 \\ & \Rightarrow x=\frac{1}{2}, x=-1 \text { (Reject) } \\ & \Rightarrow y^2=2\left(\frac{1}{2}\right) \\ & \Rightarrow y= \pm 1 \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { Area bounded } & =\int_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right) d y \\ & =\left.\left(\tan ^{-1} y-\frac{y^3}{6}\right)\right|_{-1} ^1 \\ & =\frac{\pi}{2}-\frac{1}{3} \end{aligned}$$