How does this equation hold (Secant method)? Consider we are approxinating a root by the secant method.
Then, the interation is given by $x_{n+1}=x_n - f(x_n)\frac{x_n - x_{n-1}}{f(x_n)-f(x_{n-1})}$.
In my text (Atkinson), it's written that:

After some manipulation, it's possible to show that $\alpha -x_{n+1}=(\alpha - x_n)(\alpha - x_{n+1})\frac{ -f''(\xi_n)}{2f'(\zeta_n)}$

I completely have no idea how the above equation is derived from the iteration formula.. How?
 A: There’s more than one way to obtain this expression.  Below, I’ll fill out the details outlined by Atkinson:
Suppose
$x_{n + 1}  = x_n  - \frac{{f\left( {x_n } \right)\left( {x_n  - x_{n - 1} } \right)}}{{f\left( {x_n } \right) - f\left( {x_{n - 1} } \right)}}$
and let $\alpha$ be a zero of $f$ (you neglected to mention this property of $\alpha$ in your question).
Then, 
$\alpha  - x_{n + 1}  = \alpha  - x_n  + \frac{{f\left( {x_n } \right)\left( {x_n  - x_{n - 1} } \right)}}{{f\left( {x_n } \right) - f\left( {x_{n - 1} } \right)}} = \alpha  - x_n  + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}$
where in the denominator of the last term, I’m using standard divided difference notation.  This can further be rewritten as
$\alpha  - x_n  + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}} = \frac{{\left( {\alpha  - x_n } \right)\left( {\alpha  - x_{n - 1} } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}\left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right]}}{{\left( {\alpha  - x_{n - 1} } \right)}} + \frac{{f\left( {x_n } \right)}}{{\left( {\alpha  - x_n } \right)\left( {\alpha  - x_{n - 1} } \right)}}} \right)$
Note that we can rewrite the parenthesized term in the right hand expression as
$\left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right]}}{{\left( {\alpha  - x_{n - 1} } \right)}} + \frac{{f\left( {x_n } \right)}}{{\left( {\alpha  - x_n } \right)\left( {\alpha  - x_{n - 1} } \right)}}} \right) =  - \left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right] - \frac{{f\left( \alpha  \right) - f\left( {x_n } \right)}}{{\left( {\alpha  - x_n } \right)}}}}{{\left( {x_{n - 1}  - \alpha } \right)}}} \right) =  - \left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right] - f\left[ {x_n ,\alpha } \right]}}{{\left( {x_{n - 1}  - \alpha } \right)}}} \right)$
This is the negative of a second-order divided difference that we can write as  $- f\left[ {x_{n - 1} ,x_n ,\alpha } \right]$.  Therefore,
$\alpha  - x_n  + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}} =  - \frac{{\left( {\alpha  - x_n } \right)\left( {\alpha  - x_{n - 1} } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}f\left[ {x_{n - 1} ,x_n ,\alpha } \right]$
Atkinson’s expression then follows from the mean value theorem for divided differences.  Note that in Atkinson’s expression, $\xi _n$ is some (not any) point which lies between the min and max of the elements of $x_{n - 1}$, $x_n$, and $\alpha$ and $\zeta_n$ is some (not any) point which lies between the min and max of the elements of $x_{n - 1}$ and $x_n$.
