Differentiate a restricted parametric curve and find an intersection

$dx/dt=-t(1+t^2)^{-3/2}$ and $dy/dt=3t^2$. For $t<0$, $dy/dx=-3t(1+t^2)^{3/2}$. At $t=0$ the uncancelled ratio is $0/0$; the one-sided slope of the traced curve tends to $0$ at the endpoint $(1,1)$.

Substitution gives $t^3+1=1+t^2-2$, so $t^3-t^2+2=0$. The printed calculator solution gives $t=-1$; algebraically the factorisation is $(t+1)(t^2-2t+2)=0$, and the quadratic has discriminant $-4$, so this is the only real root and satisfies $t\le0$. Hence $P=(1/\sqrt2,0)$.