My textbook, Oneill's Elementary Differential Geometry, defines a curve in terms of the function $f$: $$f(t)=\begin{cases}0, &\text{for } t\leq0\\ e^{-\frac{1}{t^2}}, &\text{for } t\gt0\end{cases}$$
The curve is defined $$a(t)={t,f(t),f(-t)}$$ The book states that the curvature is only zero at $t=0$. But when I calculate it, I get that the curvature is proportional to the length of $$a'(t)\times a''(t)$$ I get the first and second components of this product equal to zero. For the third component, for $t\gt0$ I get $$4\frac{e^{-1/t^2}}{t^6}-6\frac{e^{-1/t^2}}{t^4}$$ Setting this equal to zero, I get $$t=\sqrt{2/3}$$ So my result is at variance with the textbook. Have I made an error? If so, where?