 Solution

The parabola opens upward, with its minimum at $(0,-1)$ and zeros at $x=\pm 1$. The quartic graph is symmetric about the vertical axis, has its flat maximum at $(0,1)$, and also has zeros at $\pm 1$.

Their intersections solve $x^4+x^2-2=(x^2-1)(x^2+2)=0$, so the only real intersections are $(-1,0)$ and $(1,0)$. Between them, $1-x^4$ is the upper curve and $x^2-1$ the lower curve.