Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation
$$a(x)y'' +b(x)y' + c(x)y = 0$$
on an interval where a(x) is never zero.
Show that the Wronskian W of g and h satisfies $$a(x)W'(x) + b(x)W(x) = 0$$
Now $$W(x) = gh' - hg$$ and $$W'(x) = h''g - g''h$$ Then I get, $$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$ To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?
You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$ \begin{align*} a(x)h'' + b(x)h' &= -c(x)h \\ a(x)g'' + b(x)g' &= -c(x)g. \end{align*} $$
Substituting into your last equation, $$ \begin{align*} a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \\ &= c(x)[-gh + hg] \\ &= 0. \end{align*} $$