With the update step $\textbf x_{t+1} = \textbf x_t - H_t^{-1}\nabla f(\textbf x_t)$, where $H_t \in \mathbb R^{d\times d}$ is symmetric and satisfies $\nabla f(\textbf x_t) - \nabla f(\textbf x_{t-1}) = H_t(\textbf x_t - \textbf x_{t-1})$, it is called a Quasi-Newton method.
I have to prove that when $H_t = \nabla^2 f(\textbf x_t)$ (which is the Newton's method), the function $f$ has to be a non-degenerate quadratic of $\textbf x \in \mathbb R^d$, which is of the form: $f(\textbf x) = \dfrac 1 2 \textbf x^\top M \textbf x - \textbf q^\top x + c$
My attempt: Starting from $$\nabla f(\textbf x_t) - \nabla f(\textbf x_{t-1}) = \nabla^2 f(\textbf x_t)(\textbf x_t - \textbf x_{t-1})$$ I transformed it into: $$\textbf x_t - \nabla^2 f(\textbf x_t)^{-1}\nabla f(\textbf x_t) = \textbf x_{t-1} - \nabla^2 f(\textbf x_t)^{-1}\nabla f(\textbf x_{t-1})$$
It seems to have interesting structure, but I don't know how to proceed
Definitely not an anwser... but hope these may be useful to you
Some general derivations for quasi newton method $$ \textbf x_{t+1} = \textbf x_t - H_t^{-1}\nabla f(\textbf x_t)\\ \textbf x_{t+1} - \textbf x_t = - H_t^{-1}\nabla f(\textbf x_t)\\ -H_t(\textbf x_{t+1} - \textbf x_t) = \nabla f(\textbf x_t)\\ $$ $$ \nabla f(\textbf x_{t+1}) - \nabla f(\textbf x_{t}) = H_{t+1}(\textbf x_{t+1} - \textbf x_{t})\\ -H_{t+1}(\textbf x_{t+2} - \textbf x_{t+1}) + H_t(\textbf x_{t+1} - \textbf x_t) = H_{t+1}(\textbf x_{t+1} - \textbf x_{t})\\ H_{t+1}(\textbf x_{t+2} - \textbf x_{t}) = H_t(\textbf x_{t+1} - \textbf x_t) \\ $$ if we set $H_t=\nabla^2 f(x_{t})$ then $$ -\nabla^2 f(x_{t})(\textbf x_{t+1} - \textbf x_t) = \nabla f(\textbf x_t)\\ \nabla^2 f(x_{t+1})(\textbf x_{t+2} - \textbf x_{t}) = \nabla^2 f(x_{t})(\textbf x_{t+1} - \textbf x_t) \\ $$ Then we need to use this to show that $f(x)$ is quadratic..