Let $\gamma$ be a geodesic defined on the surface $S$ by $\gamma(t) = \sigma(u(t),v(t))$ where $\sigma$ is a surface patch. Let us assume the following for the first fundamental form: $E = 1$, $F = 0$. The first geodesic equation for such case reduces to $u'' - (v')^2 \frac{G_u}{2} = 0 $. I wonder how to prove that if additionally $G_v = 0$, then the above equation can be expressed as $(u'G)' =0$ (exercise from Tapp's book).
Differentiation gives $u''G + u'(u'G_u+v'G_v) = 0$, and so $u''G + (u')^2G_u = 0$, still I do not see how to get here. I found a similar expression to $(u'G)' =0$ in the proof of Clairaut's theorem, but the theorem is about special case of surface of evolution, polar coordinates are used, and the expression is derived from the aditional assumption of the geodesic being unit-speed which makes $\gamma''$ orthogonal to $\sigma_\theta$, so from $\langle\gamma'', \sigma_\theta\rangle = 0$ after some calculations we get $(\theta'G)' =0$ .
That is not true. For example, the flat metric is given by (under polar coordinates $(u, v) = (r, \theta)$)
$$ E =1, \ \ \ F = 0, \ \ \ G = r^2.$$
Since $(r(t), \theta(t)) = (t, \theta_0)$ is a (radial) geodesic, with $r' = 1, \theta' =0$, we see that $$ r' G = 1 \cdot (r(t))^2 = t^2$$ is not constant in $t$.
Indeed, whenever the first fundamental form is of the form $$ E =1, \ \ \ F = 0, \ \ \ G = G(u, v),$$ then $(u(t), v(t)) = (t, v_0)$ is always a geodesic (by direct checking).
What I can get is: Using that geodesics has constant length:
$$\tag{1} C = \| \gamma'\|^2 = (u')^2 + G (v')^2,$$ together with your first geodesic equation, \begin{align} C &= (u')^2 + G \frac{2u''}{G_u} \\ C G_u &= G_u (u')^2 + G 2u''\\ C G_u u'&= G_u (u')^3 + G 2u'u'' \\ (C G)' &= (G (u')^2 )' \end{align}
or that $G (C- (u')^2)$ is constant. Using (1), we have $G^2 (v')^2$, or $Gv'$, is constant.
(So it seems that either your post or the book has a typo)