Following this answer to a related question we have
$$ e_{n+1} = e_n e_{n-1} \frac{f[x_{n-1},x_n,\xi]}{f[x_{n-1},x_n]}$$
where $\xi$ is the zero and $e_k = \xi - x_k$ is the error at the $k$th iteration. We want to achieve a net reduction in the size of the error, hence our goal should be to ensure that
$$ \left | e_{n-1} \frac{f[x_{n-1},x_n,\xi]}{f[x_{n-1},x_n]} \right| \leq \lambda < 1$$
for some fixed $\lambda \in [0,1)$ so that the sequence will converge at least linearly. To that end, we shall deploy the mean value theorem for divided differences. Specifically, if $f$ is $C^2(\mathbb{R},\mathbb{R})$, there is at least one $\theta_n$ in the smallest interval containing $x_{n-1}, x_n, \xi$ such that
$$ f[x_{n-1},x_n,\xi] = \frac{1}{2} f''(\theta_n)$$ and there is at least one $\nu_n$ between $x_{n-1}$ and $x_n$ such that
$$ f[x_{n-1},x_n] = f'(\nu_n).$$
We see that in order to achieve our goal, we must gain control of the term $$\frac{1}{2} \frac{f''(\theta_n)}{f'(\nu_n)}.$$ One way to do this to first consider an arbitrary $\delta > 0$ and define $$M_\delta = \sup \left \{ \frac{1}{2} \left| \frac{f''(x)}{f'(y)}\right| \: : \: (x,y) \in [\xi-\delta,\xi+\delta]^2 \right \}. $$
It is easy to see that if $f$ is $C^2(\mathbb{R},\mathbb{R})$ and if $f'(\xi) \not = 0$, then $M_\delta$ is well-defined for $\delta$ sufficiently small and
$$ M_\delta \rightarrow \frac{1}{2} \left| \frac{f''(\xi)}{f'(\xi)} \right|, \quad \delta \rightarrow 0, \quad \delta > 0.$$
Now choose $\delta$ so small that
$$ \delta M_\delta < 1$$
and define $\lambda = \delta M_\delta$. Now choose $x_{0}, x_1 \in [\xi-\delta, \xi+\delta]$, then $|e_0| \leq \delta$ and
$$|e_2| \leq |e_1| \delta M_{\delta} \leq |e_1| \lambda < |e_1| \leq \delta.$$
We conclude that $x_2 \in [\xi-\delta, \xi+\delta]$, and this observation allows us to proceed inductively and show that the sequence does not leave the interval $[\xi-\delta, \xi+\delta]$ and $$|e_{n+1}| \leq \lambda^n |e_1|.$$
This show that the secant method converges and that the order of convergence is at least linear. We assumed that $f$ is $C^2(\mathbb{R},\mathbb{R})$, that $f'(\xi) \not = 0$, that $\delta M_\delta < 1$ and that $\xi - \delta \leq x_0, x_1 \leq \xi + \delta$ to achieve this result.
