Showing that $x_{k+2}$ is a point which approximates a maximum? Suppose that $x_k<x_{k+1}$ and $f'(x_k)>f'(x_{k+1})$. 
How can I show that the secant method will give $x_{k+2}$ as a point which approximates a maximum?
$$x_{k+2}= x_k-\frac{f'(x_k)(x_{k+1}-x_k)}{f'(x_{k+1})-f'(x_k)}$$
 A: I don't know whether you're looking for a logically rigorous proof or a heuristic argument, but if the former then you ought to say more than what you've said, so I'll boldly assume the latter, and maybe I'm wrong.
There is a reason why it's called the secant method, and the reason entails the formula for $x_{k+2}$ that you give, whereas the formula does not tell you what the reason is.
For a sufficiently well-behaved function $f'$, the secant method gives approximations to points at which $f'=0$.  Whether that implies there is a maximum or a minimum or neither, of the function $f$ whose derivative is $f'$ depends on other things, of which you've said nothing.
The idea is that to look at two points $$(x_k,f'(x_k))\quad \text{and} \quad(x_{k+1},f'(x_{k+1})) \tag 1$$ on the graph of $f'$, both of which are near a point $x$ at which $f'(x)=0$, and draw the secant line that through those two points.  The point at which the secant line crosses the $x$-axis will be taken to be $x_{k+2}$.  The way to find $x_{k+1}$ is to use the fact that the three points $(x_k,f'(x_k))$, $(x_{k+1},f'(x_{k+1}))$, and $(x_{k+2},0)$ are on a common straight line.  Since they are on a common straight line, the slope of the line through any two of them must be the say as the slope of the line through any other pair of them.  The slope of the line passing through $(x_k,f'(x_k))$ and $(x_{k+1},f'(x_{k+1}))$ is of course
$$
\frac{\Delta f(x)}{\Delta x} = \frac{f'(x_k)-f'(x_{k+1})}{x_k-x_{k+1}}. \tag 2
$$
The slope of the line through $(x_k,f'(x_k))$ and $(x_{k+2},0)$ is
$$
\frac{f(x_k)-0}{x_k-x_{k+1}}. \tag 3
$$
So set $(2)$ equal to $(3)$ and solve for $x_{k+2}$, and you've got the formula that you wrote for finding $x_{k+2}$.
The question then is: Why should $x_{k+2}$ approximate a point $x$ at which $f'(x)=0$?  And the answer is that the secant line should approximate the graph of $f'$.  One is a curve and the other is a straight line, but they both pass through the two points $(1)$, so if there isn't much curvature in that vicinity then the curve is close to a straight line and hence the point where the curve crosses the $x$-axis is close to the point where the straight line crosses the $x$-axis.
But "isn't much curvature" is an imprecise statement.  Before one can rigorously prove any results, one would have to choose among the various possible ways of making it precise.
