Let $f,g$ be smooth functions near $0$. Suppose the Wronskian determinant $$ W(f,g)(t)=f(t)g'(t)-f'(t)g(t) $$ is constantly zero.
Is it true that $W(f',g')$ is also constantly zero? Note that we are NOT assuming real-analyticity of $f,g$, so we cannot simply use linear dependence.
Let $f,g \in C^2(I,\mathbb{R})$ for some open interval $I \subseteq \mathbb{R}$ with $W(f,g)(t)=0$ $(t \in I)$. Fix any $t_0 \in I$ and for $\delta > 0$ let $J_\delta:= (t_0-\delta,t_0+\delta)$.
Case $g(t_0)\not=0$: Then $g(t) \not=0$ $(t \in J_\delta)$ for some $\delta> 0$, and $$ \left(\frac{f}{g}\right)'(t)= - \frac{W(f,g)(t)}{g(t)^2}=0 \quad (t \in J_\delta), $$ hence $f=\lambda g$ on $J_\delta$ for some $\lambda \in \mathbb{R}$. Thus $W(f',g')(t)=0$ $(t \in J_\delta)$.
Case $g(t_0)=0$: If $g(t)=0$ in some interval $J_\delta$, then again $W(f',g')(t)=0$ $(t \in J_\delta)$. Otherwise there is a sequence $(t_n)$ in $I$ with $t_n \to t_0$ and $g(t_n)\not=0$ $(n \in \mathbb{N})$. By the first case applied to $t_n$ we get $W(f',g')(t_n)=0$ $(n \in \mathbb{N})$ and since $t \mapsto W(f',g')(t)$ is continuous $n \to \infty$ yields $W(f',g')(t_0)=0$.
Thus in both cases $W(f',g')(t_0)=0$. Therefore $W(f',g')(t)=0$ $(t \in I)$.