Hint :
Take the expression you have and solve it as an ODE :
$$7f(x) + xf'(x) = 0 \Leftrightarrow \frac{f'(x)}{f(x)} = - \frac{7}{x} \Rightarrow f(x) = \frac{c_1}{x^7}$$
For $x=c$ the expressions yields :
$$f(c) = \frac{c_1}{c^7}$$
This can only be equal to $0$ iff $c_1 = 0 \Rightarrow f(x) = 0$.
For $f(x)=0$ though the hypothesis holds, since $f$ will be continuous in $[a,b]$, differentiable in $(a,b)$ and it obviously is $f(a) = f(b) = 0$. This, by Rolle's Theorem implies that $\exists c \in (a,b) : f'(c) =0$, which is also true, since if $f(x)=0$ then also $f'(x) = 0$.