Why is the following statement true:
Suppose that $f$ is twice continuously differentiable on open set $U \subset \mathbf{R}^n$. Then for all $x, y \in U$ there exists $t \in [0, 1]$ such that $z = x + t(y - x)$ and such that $$ f(y) = f(x) + \langle \nabla f(x), y- x \rangle + (1/2) \langle y - x, \nabla^2f(z)(y-x)\rangle. $$
It vaguely looks like an application of the mean value theorem, but I can't seem to show it.
General domains
The claim is false for a general open connected set $U$. For a counterexample, let $U=\mathbb{R}^2\setminus \{(x_1, 0):x_1\ge 0\}$ be a slit plane. Define $$ f(x) = \begin{cases} x_1^3\quad &\text{if }x_1,x_2>0\\ 0 &\text{otherwise} \end{cases} $$ This is a twice continuously differentiable function: e.g., the second derivative in the $x_1$ direction is $$ f(x) = \begin{cases} 6x_1\quad &\text{if }x_1,x_2>0\\ 0 &\text{otherwise} \end{cases} $$ which is continuous in $U$. The other second partials are identically zero.
Choose $x=(1, -1)$ and $y=(1, 1)$. Then the claim takes the form $$ 1 = (1/2) \langle y - x, \nabla^2f(z)(y-x)\rangle = 2 \frac{\partial^2 f}{\partial x_2^2}(z) = 0 $$ which is false.
Convex domains
The correct statement would assume that $U$ is open and convex, not merely connected. Then the proof follows by considering the function $g(t) = f(x+t(y-x))$ of a real variable $t\in [0, 1]$ and applying Taylor's theorem with the Lagrange form of the remainder.