Let $q(x)$ be the quadratic through the $(x_i, f(x_i))$. Set $g(x) = f(x) - q(x)$. Then $g(x_1) = g(x_2) = g(x_3) = 0$ and you want to show $g''(x)=0$ for some $x$. Apply Rolle's theorem to get that $g'$ has zeroes $y_1$ and $y_2$ in $(x_1, x_2)$ and $(x_2, x_3)$ respectively, then apply it again to get that $g''$ has a zero in $(y_1, y_2)$.

I can sort of give you a pointer to the literature. Rename your points $x_1$, $x_2$, ..., $x_{n}$ and $z$ with $x_1 < x_2 < \ldots < x_{n}$. Let $q(t)$ be the degree $n-1$ Lagrange interpolating polynomial for $f$ through $(x_i, f(x_i))$. Then the degree $n$ Lagrange interpolating polynomial through the $(x_i, f(x_i))$ and $(z, f(z))$ is
$$r(t) = q(t) + \frac{f(z)-q(z)}{\prod (z-x_i)} \prod (t-x_i).$$
You want to know that there is some $c \in (\min(x_1,z), \max(x_n,z))$ where $f^{(n)}(c) = r^{(n)}(c)$.
In other words, where $f^{(n)}(c) = \frac{f(z) - q(z)}{\prod (z-x_i)} n!$. You can rearrange this to give
$$f(z) = q(z) + \frac{f^{(n)}(c)}{n!} \prod (z-x_i).$$
This is a known formula for the error in Lagrange interpolation. Here is a Wikipedia article. The case where $x_1=x_2=\cdots = x_n$ is Taylor's theorem with the Lagrange form of the remainder; Tim Gowers has a nice blog post. Comments on Prof. Gowers' post suggest that Atkinson’s Intro to Numerical Analysis would give you a more official source; I haven't read it.

I don't know a source where the $n+1$ points are treating on equal footing, but my ignorance of analysis literature is no sign that it doesn't exist.