According to Differential Geometry by Pressley:
$\mathsf{Proposition}\:\mathsf{1.3}$
Any reparametrisation of a regular curve is regular.
Proof $\it{1.3}$
Suppose that $\gamma$ and $\tilde{\gamma}$ are related as in Definition 1.5, let $t=\phi(\tilde{t})$, and let $\psi=\phi^{-1}$ so that $\tilde{t}=\psi(t)$. Differentiating both sides of the equation $\phi(\psi(t))=t$ with respect to $t$ and using the chain rule gives $${{\raise{0.3pt}{d\phi}}\above{0.5pt}{d\tilde{t}}} \frac{d\psi}{\lower{1pt}{dt}}=1.$$ This shows that $d\phi/d\tilde{t}$ is never zero. Since $\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))$, another application of the chain rule gives $${{\raise{0.3pt}{d\tilde{\gamma}}}\above{0.5pt}{d\tilde{t}}}=\frac{d\gamma}{\lower{1pt}{dt}}{{\raise{0.3pt}{d\phi}}\above{0.5pt}{d\tilde{t}}},$$ which shows that $d\tilde{\gamma}/d\tilde{t}$ is never zero if $d\gamma/dt$ is never zero. $\hspace{6.2cm}\square$
Now suppose the parabola $y=x^2$ and the parametrazation $\gamma(t) = (t, t^2)$ and also reparametrazation $\delta(t)=(t^3, t^6)$. Obviously the function $\phi(t)=t^3$ is strictly monotone and continuously differentiable. Nonetheless, $\gamma(t)$ is regular but $\delta(t)$ is not! What did I do wrong? Or is the proof incorrect?
Maybe the following definition from the same book is wrong; that is, it must include that $d\phi/dt $ is never zero in all its domain?
$\mathsf{Definition}\:\mathsf{1.5}$
A parametrised curve $\tilde{\gamma}:(\tilde{\alpha},\tilde{\beta})\to\mathbf{R}^n$ is a reparametrisation of a parametrised curve $\gamma:(\alpha,\beta)\to\mathbf{R}^n$ if there is a smooth bijective map $\phi:(\tilde{\alpha},\tilde{\beta})\to(\alpha,\beta)$ (the reparametrisation map) such that the inverse map $\phi^{-1}:(\alpha,\beta)\to(\tilde{\alpha},\tilde{\beta})$ is also smooth and $$\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))\:\:\text{for all}\:\:\tilde{t}\in(\tilde{\alpha},\tilde{\beta}).$$
Best Answer
Yes, if $\phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $\frac{d\phi}{dt} \ne 0$.
This is because $\phi \circ \phi^{-1} = \text{id}$ so $$1 = \frac{d(\text{id})}{dt} = d(\phi^{-1} \circ \phi) = \frac{d\phi^{-1}}{d\phi} \frac{d\phi}{dt}$$
In particular we see that $\frac{d\phi}{dt} \ne 0$.