Show that curvature and torsion are $κ = \frac{\left|γ'∧γ''\right|}{\left|γ'\right|^3}$ and $\tau = \frac{(γ'∧γ'')\cdot γ'''}{\left|γ'∧γ''\right|^2}$

I'll do the curvature, you try the torsion.

Let $\beta:[s_0,s_1] \to \mathbb R^3$ be the reparametrization of $\gamma:[t_0,t_1] \to \mathbb R^3$ by arc length i.e: $\gamma=\beta \circ s$ where $s:[t_0,t_1]\rightarrow[s_0,s_1]$ ; $s(t)=\int_{t_o}^t||\gamma'(u)||du$.

We have:

$$\gamma' =\frac{\mathrm d\gamma}{\mathrm dt} =\frac{\mathrm d\beta}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt} =T(s) \cdot||\gamma'(t)||\tag1$$ where $T(s)$ is the tangent vector of $\gamma$. From the equation we see that $T(s)=\gamma'(t)/||\gamma'(t)||$. Then differentiating $\gamma'$ and applying the chain rule, we get:

$$\gamma'' = \frac{\mathrm d^2\gamma}{\mathrm dt^2} =\left( \frac{\mathrm dT}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt}\right) \cdot ||\gamma'(t)||+ T(s) \cdot \left( \frac{\langle\gamma'(t),\gamma''(t)\rangle}{||\gamma'(t)||}\right)\tag2$$

And since $T'=k \cdot N$, where $N$ is the normal vector of $\gamma$, we get:

$$ \gamma'' =k \cdot N \cdot ||\gamma'||^2+\left( \frac{\langle\gamma',\gamma''\rangle}{||\gamma'||}\right)\cdot T(s)\tag3$$

Computing the binormal $B$:

$$B = \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||} =T\times N\tag4$$

We get the following relation:

$$ \gamma'(t)\times \gamma''(t)=k \cdot ||\gamma'(t)||^3 \cdot T\times N\tag5$$

Finally:

$$N = B\times T = \left( \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||}\right)\times \frac{\gamma'(t)}{||\gamma'(t)||}=\frac{k \cdot ||\gamma'(t)||^3 \cdot T\times N}{||\gamma'(t)\times \gamma''(t)||}\times \frac{\gamma'(t)}{||\gamma'(t)||}\tag6$$

Since $||N||=||T\times N||=1$ and $k\geq 0$, we get: $$k=\frac {||\gamma'(t)\times \gamma''(t)||}{||\gamma'(t)||^3}.$$

Let $I=[a,b]$ and $t(s):[0,L]\to I$ be a differentiable function whit $s: [0,L]\to [a,b]$ inverse differentiable and $$ L=\int_{a}^{b} \|\gamma^\prime(t)\| \; d\, t \qquad\mbox{ and }\qquad s(t)=\int_{0}^{t} \|\gamma^\prime(t)\| \; d\, t $$ Fix the notation $\kappa(t)=\kappa(t(s))=\kappa(s)$ and $\gamma(t)=\gamma(t(s))=\gamma(s)$. We have by definition of curvature $\kappa(s)=\|\gamma^{\prime\prime}(s)\|$. Note that \begin{align} \frac{d^2}{d t^2}\gamma(s)= & \frac{d}{d t}\left[\frac{d}{dt}\gamma(s) \right] \\ = & \frac{d}{d t}\left[\gamma^{\prime}(s)\cdot \left(\frac{d s}{dt}\right)\right] \\ = & \left[ \gamma^{\prime\prime}(s)\cdot \left(\frac{d s}{dt}\right) + \gamma^{\prime}(s)\cdot \frac{d s}{dt}\left(\frac{d s}{dt}\right) \right]. \end{align} Using Frenet-Serret equations $$ \left\{ \begin{array}{rl} \gamma^\prime(s)=&T(s)\\ T^\prime(s)=& \kappa(s)\cdot N(s)\\ N^{\prime}(s)= &-\kappa(s)\cdot T(s) + \tau(s)\cdot B(s)\\ B^\prime(s)=& \tau(s)\cdot B(s) \end{array} \right. $$ we have $$ \left\{ \begin{array}{rl} \frac{d^2}{d\, t^2}\gamma(s)=& \frac{d\,s}{d\,t} T(s) \\ \frac{d^2}{d\,t^2}\gamma(s)=& \kappa(s)N(s)\left(\frac{ds}{dt}\right)+T(s)\left(\frac{d^2 s}{d\,t^2} \right) \end{array} \right. $$ Calculate the cross product below \begin{align} \gamma^{\prime}(t)\times \gamma^{\prime\prime}(t)= & \frac{d }{dt}\gamma(s)\times \frac{d^2 }{dt^2}\gamma(s) \end{align} knowing that $\{T(s), N(s), B(s)\}$ is a basis of orthonormal vectors. After replacing the equations $\frac{d\,s}{d\,t}= \|\gamma^\prime(t)\|$ and $\frac{d^2\,s}{d\,t^2}=-\frac{\|\gamma^\prime(t)\|}{|\langle \gamma^\prime,\gamma^{\prime\prime}(t)\rangle|}$. Use too the equation $$ \langle u\times v, w \rangle = det(u,v,w) \quad \forall w\in\mathbb{R}^3 $$ for simplifications. After doing some algebraic manipulations to obtain the curvature.The formula for calculating the torsion is analogous. Just obtain the torsion use the expression below. $$ \left\langle \gamma^\prime(t)\times \gamma^{\prime\prime}(t),\gamma^{\prime\prime\prime}(t) \right\rangle = \left\langle \frac{d}{d\,t}\gamma(s)\times\frac{d^2}{d\,t^2}\gamma(s) \, ,\, \frac{d^3}{d\,t^3}\gamma(s) \right\rangle $$