How to tell whether a curve has a regular parametrization?

$\newcommand{\Reals}{\mathbf{R}}$Let $I$ be a non-empty open interval of real numbers, and $\gamma:I \to \Reals^{n}$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)

At each point where $\gamma$ is regular, define the unit tangent field $$ T(t) = \frac{\gamma'(t)}{\|\gamma'(t)\|}. $$ By continuity, the zero set of $\gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $\gamma' = 0$, "excise them and push their endpoints together").

Claim: The following are equivalent:

(i) $\gamma$ has a regular $C^{1}$ parametrization.

(ii) $T$ has a continuous extension to $I$.

(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $\Gamma(t) = \gamma(\tau(t))$ for some differentiable function $\tau$ with positive derivative, then the unit tangent field of $\Gamma$ at $t$ is the unit tangent field of $\gamma$ at $\tau(t)$. If $\gamma$ has a regular $C^{1}$ reparametrization $\Gamma$, then the unit tangent field of $\Gamma$ continuously extends the unit tangent field of $T$.

(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then $$ \Gamma(s) = \int_{0}^{s} T(t)\, dt $$ is a regular parametrization of $\gamma$.

For the curve in question, we have $$ \gamma(t) = (t^{3}, t^{6}),\qquad \gamma'(t) = (3t^{2}, 6t^{5}),\qquad T(t) = \frac{(3t^{2}, 6t^{5})}{\sqrt{9t^{4} + 36t^{10}}} = \frac{(1, 2t^{3})}{\sqrt{1 + 4t^{6}}}, $$ which has a continuous extension at $t = 0$ (given by the same formula).

For a cusp, say, we have $$ \gamma(t) = (t^{2}, t^{3}),\qquad \gamma'(t) = (2t, 3t^{2}),\qquad T(t) = \frac{(2t, 3t^{2})}{\sqrt{4t^{2} + 9t^{6}}} = \frac{t(2, 3t)}{|t|\sqrt{4 + 9t^{2}}}, $$ which does not extend continuously to $0$.