if the $W(f,g) \equiv 0$ (Wronskian of Independent functions) then they both get 0 at same point let $f,g$ be differential and Independent functions at $(a,b)$.
we also know that their Wronskian is identically zero : $\forall x\in (a,b) \ \  W(f,g)(x) = 0$.
show that $\exists x_0 \in (a,b) \ s.t. \ f(x_0)= g(x_0) = 0$.
EDIT: by independent functions i mean that if  $\forall x \in (a,b) : \alpha f(x) + \beta g(x) = 0 \Rightarrow \alpha = \beta = 0$

the unimportant part:
*my try: i thought that since $\forall x \in (a,b):W(f,g)(x) = f(x)g'(x) - f'(x)g(x) = 0$ we can infer that for every point $c$ in $(a,b)$ so that $f(c) \neq 0 $ we can easily see that $(\frac{g}{f})'(c) = 0$ (and that's why for every such point c : $\exists C$ (a constant) s.t. $g(c) - Cf(c) = 0 $ ) and doing so similarly for all point c that $g(c) \neq 0$ we get that $\exists C $ (a constant) s.t. $g(c) - C f(c) = 0$ and if there isn't a point $x_0 $ such that $f(x_0) = g(x_0) = 0$ that means that f,g are dependent functions by contradiction.
that argument is not correct, because it doesn't guaranty that the constant is uniform over $(a,b)$ nor that f,g get zero value somewhere over $(a,b)$ 
 A: Suppose first that $g(x) \ne 0$ for all $x \in (a,b)$.  Then $\frac{f}{g}$ has first derivative constantly equal to zero on the whole of $(a, b)$, consequently there exists a real constant $C$ such that $f(x) = C g(x)$ for all $x \in (a,b)$, that is, $f$ and $g$ are linearly dependent.
So, there is $x_0 \in (a,b)$ with $g(x_0) = 0$.  If $f(x_0) = 0$ we are done.  Now, suppose to the contrary that for any $\xi \in (a, b)$ for which $g(\xi) = 0$ there holds $f(\xi) \ne 0$.  Fix such a $\xi$.  By continuity, there is a subinterval $(\alpha, \beta) \subset (a, b)$, $\xi \in (\alpha, \beta)$ such that $f(x) \ne 0$ for all $x \in (\alpha, \beta)$.  Repeating the reasoning from the first paragraph, but applied to $\frac{g}{f}$ on $(\alpha, \beta)$, we see that there exists a real constant $D$ such that $g(x) = D f(x)$ for all $x \in (\alpha, \beta)$, in particular, for $x = \xi$.  As $f(\xi) \ne 0$, we must have $D = 0$.  Consequently, $g$ is constantly equal to zero on $(\alpha, \beta)$.
The (nonempty) set $\{\, \xi \in (a,b): g(\xi) = 0 \,\}$ is, by the continuity of $g$, closed in the relative topology of $(a, b)$.  But we have proved that that set is open, so, as $(a, b)$ is connected, it must be equal to the whole of $(a, b)$.  Thus $g \equiv 0$ on $(a, b)$, so $f$ and $g$ are linearly dependent.
