Equivalently, let $\gamma(t)$ is a unit-speed curve with $\kappa(t) > 0$ and $\tau(t)$ never equal to zero. We claim $\gamma(t)$ is spherical iff
$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
First assume that $\gamma(t)$ of spherical. Then there is some point $a$, the center of the sphere, such that
$$||\gamma - a||^2 = r^{2}$$
for the radius of the sphere $r$. Rewrite this as
$$(*) \,\,\,\,\,\,(\gamma - a) \cdot (\gamma - a) = r^{2} $$
To get the equality above, we are going to keep on taking messier and messier derivatives. C'est la vie. You should recall important equalities for unit-speed curves, like $\dot{t} = \kappa n$ and $\dot{b} = -\tau n$. I will use these implicitly in the calculation.
To start, take the derivative of $(*)$ to get
$$ t \cdot (\gamma - a) = 0$$
Now, take the derivative again
$$t \cdot t + \kappa n \cdot (\gamma - a) = 0$$
Note that $t \cdot t = 1$, so we can rewrite this as
$$ n \cdot (\gamma - a) = -\frac{1}{\kappa}$$
Take the derivative of both sides again, to get
$$ n \cdot t + (-\kappa t + \tau b) \cdot (\gamma - a) = \frac{\dot{\kappa}}{\kappa^{2}}$$
Since $t \cdot (\gamma - a) = 0$, this reduces to
$$ b \cdot (\gamma - a) = \frac{\dot{\kappa}}{\tau\kappa^{2}}$$
Take the derivative again to get
$$b \cdot t - \tau n \cdot (\gamma - a) = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
which is easily seen to be equivalent to
$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
Now, we go the other direction, so assume
$$(*) \,\,\,\,\,\, \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
Let us define new quantities $\rho = 1/\kappa$ and $\sigma = 1/\tau$. Then $(*)$ implies that
$$(\dagger) \,\,\,\,\, \rho = \frac{d}{ds}\big(-\sigma(\dot{\rho}\sigma)\big)$$
Consider the quantity $\rho^2 + (\dot{\rho}\sigma)^2$. Let us take its derivative. By $(\dagger)$, we would have
$$\frac{d}{ds} (\rho^2 + (\dot{\rho}\sigma)^2) = 2\rho\dot{\rho} + 2(\rho\sigma)\frac{d}{ds}(\rho\sigma) = 0$$
Hence $\rho^2 + (\dot{\rho}\sigma)^2$ is some positive constant $r^2$. Let us go further and define the following "curve" $a$ (think about its definition a little)
$$a = \gamma + \rho n + \dot{\rho}\sigma b$$
Now, we take the derivative of $a(t)$, noting the equation following $(\dagger)$,
$$\dot{a} = t + \dot{\rho}n + \rho(-\kappa t + \tau b) + \frac{d}{ds} (\dot{\rho}\sigma)b + (\dot{\rho}\sigma)(-\tau n) = 0$$
and so $a$ is a constant, and is the center of our sphere by the following calculation
$$\parallel \gamma - a \parallel^2 = ||\rho n + \dot{\rho}\sigma b||^2 = \rho^2 + (\dot{\rho}\sigma)^2 = r^2$$
and the radius of this sphere is $r$. This demonstrates that $\gamma(t)$ lies on the surface of a sphere of radius $r$, and hence is spherical. QED.
Here are a couple of explicit hints. For a theoretical problem, you (most) always want to assume arclength parametrization and avoid those cross-product expressions. I will use $T,N,B$ for the Frenet frame. Differentiation is with respect to arclength.
(1) If $\gamma$ is a curve on the unit sphere, show that $\gamma\cdot N = -1/\kappa$.
(2) By definition (since the unit normal to the sphere is the position vector), $\kappa N = \kappa_g (\gamma\times T) + \kappa_n\gamma$, and $\kappa_n = -1$.
First, this tells you that $\kappa = \sqrt{\kappa_g^2+1}$.
(3) Next, differentiate this and precisely one term will involve $\tau B$. We can isolate $\tau$ by dotting with $B$. You should then solve for $\tau$ in terms of $\kappa_g'$ and $\kappa$.
Best Answer
Let $\omega$ denote a period for the curve $\vec{r}$. It is useful to know that for a curve $\vec{r}$ lying on $S^{2}\subset\mathbb{E}^{3}$, the curvature $\kappa$ and torsion $\tau$ can be expressed in terms of the function $$ j(s):=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right).\tag{*} $$ Namely, I claim that \begin{align} &\kappa(s)=\sqrt{1+j(s)^{2}},\tag{1}\\ &\tau(s)=\frac{j'(s)}{1+j(s)^{2}}\tag{2}. \end{align} Assuming for now that the equalities $(1)$ and $(2)$ are true, we get $$ \int_{0}^{\omega}\frac{\tau(s)}{\kappa(s)}ds=\int_{0}^{\omega}\frac{j'(s)}{(1+j(s)^{2})^{3/2}}ds=\int_{j(0)}^{j(\omega)}\frac{1}{(1+u^{2})^{3/2}}du,\tag{3} $$ substituting $u:=j(s)$ in the last equality. Now, since $\vec{r}(s)=\vec{r}(s+\omega)$, we know in particular that $\vec{r}(0)=\vec{r}(\omega)$, $\dot{\vec{r}}(0)=\dot{\vec{r}}(\omega)$ and $\ddot{\vec{r}}(0)=\ddot{\vec{r}}(\omega)$. So from $(*)$ it is then clear that $j(0)=j(\omega)$, which implies that the integral $(3)$ is indeed zero.
Now it remains to prove the equalities $(1)$ and $(2)$.
(1) Since $\vec{r}(s)$ lies on the unit sphere and has speed $1$, we have that $\{\vec{r}(s),\dot{\vec{r}}(s),\vec{r}(s)\times\dot{\vec{r}}(s)\}$ is an orthonormal frame. Expressing $\ddot{\vec{r}}(s)$ in this frame gives \begin{align*} \ddot{\vec{r}}(s)&=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+(\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s))\dot{\vec{r}}(s)+(\ddot{\vec{r}}(s)\cdot \vec{r}(s)\times\dot{\vec{r}}(s))\vec{r}(s)\times\dot{\vec{r}}(s)\\ &=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s)\\ &=-\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s).\tag{4} \end{align*} Here we use that $\vec{r}(s)\cdot\vec{r}(s)=1$, so that $\dot{\vec{r}}(s)\cdot\vec{r}(s)=0$ and therefore $\ddot{\vec{r}}(s)\cdot\vec{r}(s)=-\dot{\vec{r}}(s)\cdot\dot{\vec{r}(s)}=-1$. We also use that $\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=0$ since $\dot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=1$. Now finally, since $\kappa(s)=\|\ddot{\vec{r}}(s)\|$, the expression $(4)$ gives $$ \kappa(s)=\|\ddot{\vec{r}}(s)\|=\sqrt{1+j(s)^{2}}. $$
(2) We can rewrite the expression (*) as follows: \begin{align*} j(s)&=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\dot{\vec{r}}(s)\times\ddot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\vec{t}(s)\times\kappa(s)\vec{n}(s)\right)\\&=\kappa(s)\vec{r}(s)\cdot\vec{b}(s). \end{align*} Differentiating the equality $\vec{r}(s)\cdot\vec{r}(s)=1$ repeatedly and using the Frenet-Serret formulas, we get \begin{align*} \vec{r}(s)\cdot\vec{t}(s)=0 &\Rightarrow \vec{t}(s)\cdot\vec{t}(s)+\kappa(s)\vec{r}(s)\cdot\vec{n}(s)=0\\ &\Rightarrow \vec{r}(s)\cdot\vec{n}(s)=-1/\kappa(s). \end{align*} So, expressing $\vec{r}(s)$ in the orthonormal frame $\{\vec{t}(s),\vec{n}(s),\vec{b}(s)\}$ gives \begin{align*} \vec{r}(s)&=(\vec{r}(s)\cdot\vec{t}(s))\vec{t}(s)+(\vec{r}(s)\cdot\vec{n}(s))\vec{n}(s)+(\vec{r}(s)\cdot\vec{b}(s))\vec{b}(s)\\ &=\frac{-1}{\kappa(s)}\vec{n}(s)+\frac{j(s)}{\kappa(s)}\vec{b}(s). \end{align*} Differentiating this equality and using the Frenet-Serret formulas gives $$ \vec{t}(s)=-\left(\frac{1}{\kappa(s)}\right)'\vec{n}(s)-\frac{1}{\kappa(s)}(-\kappa(s)\vec{t}(s)+\tau(s)\vec{b}(s))+\left(\frac{j(s)}{\kappa(s)}\right)'\vec{b}(s)-\frac{j(s)}{\kappa(s)}\tau(s)\vec{n}(s)\tag{5} $$ So the coefficient of $\vec{b}(s)$ in $(5)$ has to be zero, which gives $$ \tau(s)=\kappa(s)\left(\frac{j(s)}{\kappa(s)}\right)'.\tag{6} $$ But we know from $(1)$ that $\kappa(s)=\sqrt{1+j(s)^{2}}$. So $$ \left(\frac{j(s)}{\kappa(s)}\right)'=\frac{j'(s)\sqrt{1+j(s)^{2}}-\frac{j(s)^{2}j'(s)}{\sqrt{1+j(s)^{2}}}}{1+j(s)^{2}}=\frac{j'(s)(1+j(s)^{2})-j(s)^{2}j'(s)}{(1+j(s)^{2})^{3/2}}=\frac{j'(s)}{(1+j(s)^{2})^{3/2}}.\tag{7} $$ So we conclude from $(6)$, $(7)$ and $(1)$ that $$ \tau(s)=\frac{j'(s)}{1+j(s)^{2}}. $$