I'm going to use the "dot" notation for derivatives with respect to $s$, so that $\dot{\mathbf r}(s) = \mathbf r'(s)$, etc. Then, assuming that the "natural parameter" $s$ is the arc-length along $\mathbf r(s)$, we have:
Since $\mathbf r(s)$ is contained in the unit sphere,
$\mathbf r(s) \cdot \mathbf r(s) = 1, \tag{1}$
from which it readily follows upon differentiation with respect to the arc-length parameter $s$, that
$\mathbf r(s) \cdot \dot{\mathbf r}(s) = 0; \tag{2}$
furthermore we see that
$\dot{\mathbf r}(s) \cdot \dot{\mathbf r}(s) = 1 \tag{3}$
by virtue of the fact that $s$ is the arc-length along $\mathbf r(s)$; $\dot {\mathbf r}(s) = \mathbf T(s)$ is the unit tangent to $\mathbf r(s)$, and a member of the Frenet-Serret apparatus (or frame) of $\mathbf r(s)$. It then follows from (1)-(3) and the elementary properties of the vector cross product that $\mathbf r(s) \times \dot{\mathbf r}(s)$ is itself a unit vector and is orthogonal to both the unit vectors $\mathbf r(s)$ and $\dot{\mathbf r}(s)$, i.e.,
$(\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)) = 1, \tag{4}$
$(\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot \mathbf r(s) = (\mathbf r(s) \times \dot{\mathbf r}(s)) \cdot \dot{\mathbf r}(s) = 0. \tag{5}$
(1)-(5) show that $\mathbf r(s)$, $\dot{\mathbf r}(s)$, and $\mathbf r(s) \times \dot{\mathbf r}(s)$ themselves form an orthonormal frame at each point of $\mathbf r(s)$; we expand the Frenet-Serret normal
$\mathbf N(s) = (1 / \kappa)\dot{\mathbf T}(s) = (1 / \kappa) \ddot {\mathbf r}(s) \tag{6}$
in terms of this frame, starting by noting that differentiation of (2) yields
$\dot{\mathbf r}(s) \cdot \dot{\mathbf r}(s) + \mathbf r(s) \cdot \ddot {\mathbf r}(s) = 0, \tag{7}$
which by virtue of (3) and (6) reads
$1 + \kappa \mathbf N(s) \cdot \mathbf r(s) = 0. \tag{8}$
(7) also has the advantage of showing that $\ddot {\mathbf r}(s) \ne 0$, so that $\kappa = \Vert \ddot {\mathbf r}(s) \Vert \ne 0$ and $\mathbf N(s)$ is well-defined at all points of $\mathbf r(s)$. (8) shows that the component of $\mathbf N(s)$ along the unit vector $\mathbf r(s)$ is in fact
$\mathbf N(s) \cdot \mathbf r(s) = -\dfrac{1}{\kappa}. \tag{9}$
The $\dot{\mathbf r}(s)$ component of $\mathbf N(s)$ vanishes for all $s$, since
$\mathbf N(s) \cdot \dot{\mathbf r}(s) = \mathbf N(s) \cdot \mathbf T(s) = 0, \tag{10}$
and finally the component of $\mathbf N(s)$ along $\mathbf r(s) \times \dot{\mathbf r}(s)$ is
$\mathbf N(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)) = \dfrac{1}{\kappa}\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)), \tag{11}$
so that $\mathbf N(s)$ may in fact be written
$\mathbf N(s) = -\dfrac{1}{\kappa} \mathbf r(s) + \dfrac{1}{\kappa}\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s))(\mathbf r(s) \times \dot{\mathbf r}(s)), \tag{12}$
and using (12) to compute $\mathbf N(s) \cdot \mathbf N(s) = 1$ yields
$\dfrac{1}{\kappa^2}(1 + (\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)))^2 = 1, \tag{13}$
which is easily re-arranged:
$\kappa = \sqrt{1 + (\ddot{\mathbf r}(s) \cdot (\mathbf r(s) \times \dot{\mathbf r}(s)))^2}, \tag{14}$
the requisite result. QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
You want to show that the torsion is zero.
Note that If $\mathbf{r}$ is the unit radius vector to the curve then
$$\mathbf{t}=\frac{d \mathbf{r}}{ds}$$
Now $\mathbf{t} \cdot \mathbf{r}=0$ so on differentiating we get
$$\frac{d \mathbf{t}}{ds}\mathbf{r}+\mathbf{t}\frac{d \mathbf{r}}{ds}=0$$
or $$\kappa \mathbf{n}\cdot \mathbf{r}+ \mathbf{t}\cdot \mathbf{t}=0$$
(Note that $\kappa\neq 0$, since otherwise we have a straight line.)
So $$\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$$ differentiating this and using that $\kappa$ is constant we get
$$(-\kappa \mathbf{t}+\tau \mathbf{b})\cdot \mathbf{r}+\mathbf{n}\cdot \mathbf{t}=0$$
which simplifies to
$$\tau \mathbf{b}\cdot \mathbf{r}=0$$
If $\tau=0$ we are done so assume
$$\mathbf{b}\cdot \mathbf{r}=0$$
and differentiate
which gives
$$-\tau \mathbf{n}\cdot \mathbf{r}+\mathbf{b}\cdot \mathbf{t}=0$$
$$-\tau \mathbf{n}\cdot \mathbf{r}=0$$
But $\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$ so again $\tau=0$.
Best Answer
Your equation is incorrect, the correct condition is
$$\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}$$
I will only show the $\Leftarrow$ part here.
Let $s$ be the arc length parametrization and $\vec{t}(s), \vec{n}(s), \vec{b}(s)$ be the vectors appear in Frenet Serret equations. Define $$\vec{\beta}(s) = \vec{\alpha}(s) + \frac{1}{\kappa(s)}\vec{n}(s) - \frac{\dot{\kappa}(s)}{\tau(s)\kappa(s)^2} \vec{b}(s)\tag{*1}$$ Differentiate it with respect to $s$, we get:
$$\begin{align} \frac{d}{ds}\vec{\beta}(s) = & \vec{t} - \frac{\dot{\kappa}}{\kappa^2}\vec{n} + \frac{1}{\kappa}(-\kappa \vec{t} + \tau \vec{b} ) - \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\vec{b} -\frac{\dot{\kappa}}{\tau\kappa^2}(-\tau\vec{n})\\ = & \left(\frac{\tau}{\kappa} -\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\right) \vec{b}\\ = & \frac{\tau\kappa^2}{\dot{\kappa}}\left(\frac{\dot{\kappa}}{\kappa^3} - \frac{\dot{\kappa}}{\tau\kappa^2}\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right) \right) \vec{b}\\ = & -\frac{\tau\kappa^2}{2\dot{\kappa}}\frac{d}{ds}\left(\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2\right) \vec{b}\\ = & \vec{0} \end{align}$$ This implies $\vec{\beta}(s) = \vec{\beta}(0)$ is a constant. From this, we get
$$ \vec{\alpha} - \vec{\beta}(0) = -\frac{1}{\kappa}\vec{n} + \frac{\dot{\kappa}}{\tau\kappa^2} \vec{b} \quad\implies\quad \left|\vec{\alpha} - \vec{\beta}(0)\right|^2 = \frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}. $$ i.e $\vec{\alpha}(s)$ lies on a sphere with $\beta(0)$ as center.
Motivation of above proof
You may wonder how can anyone figure out the magic formula in $(*1)$. If you work out the $\Rightarrow$ part of the proof where $\alpha(s)$ lies on a sphere centered at $\vec{c}$, you should obtain a bunch of dot products between $\vec{\alpha}(s) - \vec{c}$ and $\vec{t}(s)$, $\vec{n}(s)$ and $\vec{b}(s)$. In particular, you should get: $$\begin{cases} \vec{t} \cdot (\vec{\alpha} - \vec{c}) & = 0\\ \vec{n} \cdot (\vec{\alpha} - \vec{c}) & = -\frac{1}{\kappa}\\ \vec{b} \cdot (\vec{\alpha} - \vec{c}) &= \frac{\dot{\kappa}}{\tau\kappa^2} \end{cases}$$
Using these, you can express the center $\vec{c}$ in terms of $\kappa, \tau$ like what we have in $(*1)$. If the curve does lie on a sphere, then the "center" should not move as $s$ changes. The proof of the $\Leftarrow$ part above is really using the given condition to verify the "center" so defined doesn't move.