Suppose $(x_k)$ is produced by the secant method applied to $f(x)$. We obtain the following identity using the mean value theorem
(a) $ξ − x_{k+1} = ξ − x_k −[f'(η_k)/f'(θ_k) ](ξ − x_k)$ for where $η_k$ is between $xk$ and $ξ$, and θk is between $x_k$ and $x_{k−1}$.
(b) $ξ − x_{k+1} = ξ − x_k +[f'(η_k)/f''(θ_k) ](ξ − x_k)$ for where $ηk$ is between $x_k$ and $ξ$, and θk is between $x_k$ and $x_{k−1}$.
(c) $ξ − x_{k+1} = − (ξ − x_k)^2 f ''(η_k) /2f ' (x_k)$ for where $η_k$ is between $x_k$ and $ξ$.
(d) $ξ − x_{k+1} = ξ − x_k + [f ' (η_k) /f ' (θ_k) ](ξ − x_k)^3$ for where $η_k$ is between $x_k$ and $ξ$, and $θ_k$ is between $x_k$ and $x_{k−1}$.
As $$ x_{k+1}=x_k-f(x_k)\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})} $$ you get $$ ξ-x_{k+1}=ξ-x_k-(ξ-x_k)\frac{f(ξ)-f(x_k)}{ξ-x_k}\frac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})} $$ which should help you to apply the mean value theorem in the relevant places.