I assume that $f$ is continuous on $[a,b]$ and twice differentiable on $(a,b)$. Let $ P(x):=f(x)-\left(f(a)\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+f(b)\dfrac{(x-a)(x-c)}{(b-a)(b-c)}+f(c)\dfrac{(x-a)(x-b)}{(c-a)(c-b)}\right). $ Then, apply Rolle's Theorem successively to $P$ and P'.
[In more detail: $P$ is continuous on $[a,b]$, twice differentiable on $(a,b)$, and $P(a)=P(c)=P(b)=0$. Therefore, by Rolle's Theorem, there exist $d$ in $(a,c)$ and $e$ in $(c,b)$ such that P'(d)=P'(e)=0. Since P' is differentiable on $(a,b)$, it is continuous on $[d,e]$, and it is differentiable on $(d,e)$. So, we may apply Rolle's Theorem to P' to find $\xi$ in $(d,e)$ with P''(\xi)=0. This $\xi$ is then as desired.]