$y_1(t) = t^2$ and $y_2(t) = t|t|$ linearly independent on [-1,1] For $t<0$, we have $t|t|= -t^2$. For $t>0$, we have $t|t|=t^2$.
Hence the Wronskian $$W[y_1,y_2](t) = y_1 y_2' -y_1' y_2  \\
= t^2 \cdot (-2t) - (2t)\cdot(-t^2) =  0 \quad t<0 \\ t^2\cdot (2t)-(2t)\cdot t^2 = 0 \quad t>0\ .
$$
In either case, the Wronskian is zero. How are these functions then linearly independent on [-1,1]?
Context: this is problem 2.1.11 of Braun's Differential Equations and Their Applications, 4th edition.

(a) Show that $y_1$ and $y_2$ are linearly dependent on the interval $0\leq t \leq 1$.


(b) Show that $y_1$ and $y_2$ are linearly independent on the interval $-1\leq t \leq 1$.

 A: You don't need the Wronskian to know these two functions are linearly independent on the interval $[-1,1]$ because two functions are linearly dependent just when one is a constant multiple of the other. That's clearly not the case here. These function are dependent on $[0,1]$ since they are identical. They are also dependent on $[-1,0]$.
At https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/3%3A_Second_Order_Linear_Differential_Equations/3.6%3A_Linear_Independence_and_the_Wronskian you can read that

Let $f$ and $g$ be differentiable on $[a,b]$. If Wronskian $W(f,g)(t_0)$ is
nonzero for some $t_0$ in $ [a,b]$ then $f$ and $g$ are linearly independent on
$[a,b]$.

Note the implication there: if the Wronskian is nonzero then the functions are independent. You are invoking the converse. But nowhere does it say that an identically zero Wronskian implies dependence.
Your example shows that the converse is in fact false. You can find it in the wikipedia page that @ChristianBlatter notes in his comment.
A: The Wronskian test seems inconclusive in this context. As you have pointed  out, $t\lvert t \rvert = -t^{2}$ for $t < 0 $ and $t \lvert t \rvert = t^{2} $ for $t \ge 0$  ( "$\ge$" since the absolute value has $0$ inclusive) so this translates to the system of equations
\begin{align*}
c_{1} t^{2} + c_{2}t^{2} &= 0 \quad ,t \ge 0  \\
c_{1}t^{2} - c_{2}t^{2} &= 0 \quad ,t <0
\end{align*}
Solving this system of equation yields
$$ 
\frac{t^{2}}{t^{2}}\begin{pmatrix}
1 & 1 & | \ \ \ 0 \\ 1 & -1  &| \ \  \ 0
\end{pmatrix} \to \begin{pmatrix} 1 & 0 & |\ \ \  0  \\ 0 & 1 & |\ \ \  0 \end{pmatrix} \implies c_{1} = c_{2} \equiv 0,$$
the system is linearly independent for arbitrary $t$ so it is linearly independent for $t \in (0,1)$.
