For a regular smooth parametrized curve $\gamma:I\rightarrow \mathbb{R}^3$ of unit speed i.e., with $||\gamma'(s)||=1$ for all $s\in I$ we define curvature of $\gamma$ at $s$ to be $||\gamma''(s)||$.
Suppose $\gamma$ is not of unit speed, there exists a parametrization $\widetilde{\gamma}=\gamma\circ \varphi : J\rightarrow I\rightarrow \mathbb{R}^3$ such that $||\widetilde{\gamma}(s)||=1$ for all $s\in J$. We then have curvature of $\widetilde{\gamma}$ at $s$ to be $||\widetilde{\gamma}''(s)||$.
Let $\gamma(s)=(x_1(s),x_2(s),x_3(s))$ then $\gamma'(s)=(x_1'(s),x_2'(s),x_3'(s))$ and $\gamma''(s)=(x_1''(s),x_2''(s),x_3''(s))$.
I am trying to define curvature of $\gamma$ using curvature of $\widetilde{\gamma}$.
Let $t\in I$ then, it is natural to define curvature of $\gamma$ at $t$ to be curvature of $\widetilde{\gamma}$ at $\varphi^{-1}(t)$ which we expect to be $$\frac{||\gamma'(t)\times\gamma''(t)||}{||\gamma'(t)||^3}.$$
We have $\widetilde{\gamma}=\gamma\circ \varphi$. So, $\widetilde{\gamma}'(\varphi^{-1}(t))=\gamma'(\varphi(\varphi^{-1}(t)))\varphi'(\varphi^{-1}(t))=\gamma'(t)\varphi'(\varphi^{-1}(t))$.
Just to get rid of those inverses, we set $\phi=\varphi^{-1}$ and we have $$\widetilde{\gamma}'(\phi(t))=\gamma'(t)\varphi'(\phi(t))=(x_1'(t)\varphi'(\phi(t)),x_2'(t)\varphi'(\phi(t)),x_3'(t)\varphi'(\phi(t))).$$ Computing its derivative, we have $$\widetilde{\gamma}''(\phi(t))=((x_1'(t)\varphi'(\phi(t)))',(x_2'(t)\varphi'(\phi(t)))',(x_3'(t)\varphi'(\phi(t)))').$$
I have computed $\gamma'(t)\times \gamma''(t)$ as $$(x_2'(t)x_3''(t)-x_2''(t)x_3'(t),x_3'(t)x_1''(t)-x_3''(t)x_1'(t),x_1'(t)x_2''(t)-x_1''(t)x_1'(t)).$$
I do not see that this is going anywhere.
Any help is welcome.
Note that $\phi(t) = \int_0^s \|\gamma'(t)\|\,dt$. In your notation, we must have $\gamma = \tilde \gamma \circ \phi$. Applying the chain rule, we find that $$ \gamma'(t) = \phi'(t) \cdot \tilde \gamma'(\phi(t))\\ \gamma''(t) = \phi''(t) \cdot \tilde \gamma'(\phi(t)) + [\phi'(t)]^2 \cdot \tilde \gamma''(\phi(t)) $$ Note, however, that $\tilde \gamma'$ and $\tilde \gamma''$ are mutually orthogonal (and yield the tangent and normal unit vectors respectively). Verify the cross product formula, noting that $\tilde \gamma ' \times \tilde \gamma ' = 0$, and $\|\tilde \gamma' \times \tilde \gamma ''\| = 1$.