 Solution

The curve is a hyperbola with two branches, vertices $(0,1)$ and $(0,-1)$, and asymptotes $y=\pm x$. It is symmetric under reflection in either coordinate axis. The requested sketch marks both vertices:

Parametrize the branches by $x=\sinh t$, $y=\pm \cosh t$, for $t\in \mathbb{R}$. The squared speed is ${\dot{x}}^2+{\dot{y}}^2={\cosh }^{2}t+{\sinh }^{2}t=\cosh 2t$, and $|\dot{x}\ddot{y}-\ddot{x}\dot{y}|=|{\cosh }^{2}t-{\sinh }^{2}t|=1$. The reciprocal of the curvature of a plane curve is thereforeSince $\cosh 2t\ge 1$, with equality only at $t=0$, the least radius is $1$, attained at $(0,1)$ and $(0,-1)$.