Draw solutions of $y'=2x\sqrt{(1-y^2)}$
Here are the slope field and some solutions:
Why this was expected? Because $2x\sqrt{(1-y^2)}>0$ for all $x>0$ and $-1<y<1$, so all solutions must be strictly increasing there. Also $2x\sqrt{(1-y^2)}<0$ for all $x<0$ and $-1<y<1$, so all solutions must be strictly decreasing there.