Existence of $x_0$ s.t. $\frac{f''( x_0 )}{2}=\frac{f( a )}{( a-b ) ( a-c )}+\frac{f( b )}{( b-a ) ( b-c )}+\frac{f( c )}{( c-a ) ( c-b )}$ 
Let $f$ be continuous on $[a,b]$, and $f''(x)$ exists on $(a,b)$. Prove that $\forall c\in(a,b), \exists x_0\in(a,b)$, s.t.
$$
\frac{f''\left( x_0 \right)}{2}=\frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}
$$

My Approach: Assume $$\dfrac{K}{2} = \frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}$$
Let
$$
g\left( x \right) =\frac{K}{2}\left( \left( a-x \right) \left( x-c \right) \left( c-a \right) \right) +f\left( a \right) \left( x-c \right) +f\left( x \right) \left( c-a \right) +f\left( c \right) \left( a-x \right) 
$$
Then $g(a)=g(c)=g(b)=0$.
Since $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exist $x_1\in(a,c),x_2\in(c,b)$ s.t. $g'(x_1)=g'(x_2)=0$.
And since $g'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $x\in(x_1,x_2)\subset (a,b)$, s.t. $g''(x_0)=0$.
From
$$
g''\left( x_0 \right) =\frac{K}{2}\cdot 2\left( a-c \right) +f''\left( x_0 \right) \left( c-a \right) 
$$
We have $K=f''(x_0)$.
My Question is, is the proof right? And can we generalize the original claim(that is, change $f''(x)$ to $f^{(k)}(x)$)?
Other proofs of the original claim is also welcomed.
 A: Your proof is (now) correct. This is related to polynomial interpolation:
$$
 p(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)}f(a) + \frac{(x-a)(x-c)}{(b-a)(b-c)}f(b) + \frac{(x-a)(x-b)}{(c-a)(c-b)}f(c)  
$$
is the Lagrange polynomial for the data points $(a, f(a))$, $(b, f(b))$, $(c, f(c))$, that is the (unique) second-degree polynomial with the property
$$ 
 p(x) = f(x) \text{ for } x= a, b, c \, .
$$
Repeated application of Rolle's theorem shows that $(p-f)''(x_0) = 0$ for some $x_0$, i.e.
$$
 \frac 12 f''(x_0) = \frac 12 p''(x_0) = \frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)} \, .
$$
Your function $g$ is essentially $f - p$ (up to some constant factor), so the above reasoning is equivalent to your proof. Writing it as an interpolation problem has the advantage that the generalization  to higher derivative becomes more apparent:

If $I \subset \Bbb R$ is an interval, $a_0, \ldots, a_n \in I$ are pairwise distinct, and $f: I \to \Bbb R$ is $n$-times differentiable, then
$$
\frac 1{n!} f^{(n)}(x_0) = \sum_{k=0}^n \frac{f(a_k)}{\prod_{j \ne k} (a_k-a_j)}
$$
for some $x_0 \in I$.

The proof is the same: We define $p$ as the Lagrange interpolation polynomial which interpolates $f$ at the $n+1$ points $x=a_0, \ldots, a_n$. $p$ has degree (at most) $n$ and $(p-f)^{(n)}(x_0) = 0$ for some $x_0 \in I$.
A: Your formula can be written as
$$\frac{\left | \begin{matrix} 1& 1& 1\\ a  & b & c \\ f(a) & f(b) & f(c) \end{matrix} \right|}{
\left| \begin{matrix} 1& 1& 1\\ a  & b & c \\ g(a) & g(b) & g(c) \end{matrix} \right|}=\frac{f''(\xi)}{g''(\xi)}
$$
(with $g(x) = x^2$).
Let $K$ be the value of the quotient. Then the function
$$F(x) =\left | \begin{matrix} 1& 1& 1\\ x  & b & c \\ f(x) & f(b) & f(c) \end{matrix} \right| - K \cdot \left | \begin{matrix} 1& 1& 1\\ x  & b & c \\ g(x) & g(b) & g(c) \end{matrix} \right|$$
is $0$ at $x=a$. Clearly it is also $0$ at $x=b$, and $x=c$. So it has $3$ distinct zeroes. Applying Rolle twice we find $\xi$ such that $F^{(2)}(\xi)=0$.
It is easy to generalize this to $n+1$ distinct $a_i$'s. One can also pass to the limit in these formulas, grouping several $a_i$ together. In particular, if $n$ of them are "equal", we get the quotient of Taylor series approximations.
