Prove there exist two distinct points $\eta,\xi \in (a,b)$ such that $f'(\eta)f'(\xi)=\left[\frac{f(b)-f(a)}{b-a}\right]^2$. 
Suppose $f(x)$ is continuos over $[a,b]$ and differentiable over
$(a,b)$. Prove there exist two distinct points $\eta,\xi \in
 (a,b)$ such that
$f'(\eta)f'(\xi)=\left[\dfrac{f(b)-f(a)}{b-a}\right]^2$.

For this purpose, if we can prove that, there exists $c\in(a,b)$ such that $$\frac{f(c)-f(a)}{c-a}\cdot\frac{f(b)-f(c)}{b-c}=\left[\frac{f(b)-f(a)}{b-a}\right]^2,\tag{*}$$then applying Lagrange MVT over $(a,c)$ and $(c,b)$ respectively, the conclusion is followed. But $(*)$ seems not to hold necessarily.
 A: There is an already an excellent and simple answer, so here is another approach which needs a little more effort.
By mean value theorem we have a $c\in(a, b) $ such that $$k=\frac{f(b) - f(a)} {b-a} =f'(c) $$ If $k=0$ then the problem is trivial so let us assume $k\neq 0$. If $f'$ equals $k$ at two distinct points then we are done. Otherwise there are values of $f'$ which are less than $k$ as well as greater than $k$ (this is a subtle point as comments on this answer indicate, see below the fold for proper justification). By intermediate value property of derivatives we can ensure that there are two numbers $l, m$ in range of $f'$ of sign same as that of $k$ such that $l<k<m$.
Again by intermediate value property the entire interval $[l, m] $ is a subset of the range of $f'$. Can you now show that there are two distinct numbers $p, q$ in this interval such that $pq=k^2$?
And this clearly generalizes to the existence of $n$ distinct points $p_1,p_2,\dots,p_n$ such that their product equals $k^n$. 

Let's come back to the point where I drew a conclusion that $f'$ takes values less than $k$ as well as greater than $k$. 
Let's assume that $c\in(a, b) $ is the only solution to $f'(x) =k$ and if $x\neq c$ then $f'(x) <k$. Then we have
\begin{align}
f(b) - f(a) &=f(b) - f(c) +f(c) - f(a)\notag\\
&=f'(x_1)(b-c)+f'(x_2)(c-a)\notag\\
&<k(b-c)+k(c-a)\notag\\
&=k(b-a)\notag
\end{align}
which is an obvious contradiction. Similarly we can handle the case when $x\neq c$ implies $f'(x) >k$.
It thus follows that if $f'(x) =k$ has only one solution in $(a, b) $ then derivative $f'$ takes values which are less than as well as greater than $k$. 
A: Construct function
$$
g(x) = [f(x) - f(a)]^2 - (x-a)^2 \left[\frac{f(b)- f(a)}{b-a}\right]^2,
$$
then we have $g(a) = g(b) = 0$. Hence, from Lagrange's MVT there exists $\xi \in (a, b)$ such that $g'(\xi) = 0$, i.e., 
$$
[f(\xi) - f(a)]f'(\xi) - (\xi - a) \left[\frac{f(b)- f(a)}{b-a}\right]^2 = 0.
$$
Therefore,
$$
\frac{f(\xi) - f(a)}{\xi - a} \cdot f'(\xi) -  \left[\frac{f(b)- f(a)}{b-a}\right]^2 = 0.
$$
Again, applying Lagrange's MVT, we can find $\eta \in (a, \xi)$ such that
$$
\frac{f(\xi) - f(a)}{\xi - a} = f'(\eta),
$$
which finishes the proof.

Remark
Following the same derivation and using mathematical induction, one can easily generalize the conclusion to any $n \geq 1$:
There exist $n$ distinct points $\xi_1, \dots, \xi_n \in (a, b)$ such that
$$
\prod_{i = 1}^n f'(\xi_i) = \left[\frac{f(b) - f(a)}{b - a}\right]^n.
$$
