This seems to be exercise 7.2.26 from the book Elementary Real
Analysis by Brian S.
Thomson,Judith B. Bruckner,Andrew M. Bruckner;
p.278.
This exercise is contained in Section 7.2.3 The Derivative as a
Magnification, where authors provide the following motivation for the
derivative at a point.
If $J$ is a sufficiently small interval having $x_0$ as an endpoint, then the ratio
$|f(J)|/|J|$ is approximately $|f'(x_0)|$, the approximation becoming "exact in
the limit." Thus $|f'(x_0)|$ can be viewed as a "magnification factor" of small
intervals containing the point $x_0$.
With this motivation we can formalize the above statement as: For very
$\varepsilon>0$ and there exists $\delta>0$ such that for any interval
$J$, such that $|J|<\delta$ and $J$ has $x_0$ as the endpoint, the inequality
$$\left|\frac{|f(J)|}{|J|}-|f'(x_0)|\right|<\varepsilon$$
holds.
By the definition of derivative you have $\delta>0$ such that
$|x-x_0|\le\delta$ implies
$|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)|<\varepsilon$.
But this also implies
$$\left|\frac{|f(x)-f(x_0)|}{|x-x_0|}-|f'(x_0)|\right|<\varepsilon.$$
If $J$ is interval of the form $[x_0,x]$ with $x_0-x<\delta$ and $y\in J$ then
$$\left|\frac{|f(y)-f(x_0)|}{y-x_0}-|f'(x_0)|\right|<\varepsilon\\
\left||f(y)-f(x_0)|-|f'(x_0)|(y-x_0)\right|<\varepsilon(y-x_0)\\
|f'(x_0)|(y-x_0)-\varepsilon(y-x_0) < |f(y)-f(x_0)| < |f'(x_0)|(y-x_0)+\varepsilon(y-x_0)\\
f(x_0)-|f'(x_0)|(y-x_0)-\varepsilon(y-x_0) < f(y) < f(x_0)+|f'(x_0)|(y-x_0)+\varepsilon(y-x_0)
$$
Considering the rightmost point of the interval (which fulfills $x-x_0=|J|$) we see that
$$|f(J)| \ge |f(x)-f(x_0)| \ge |f'(x_0)||J|-\varepsilon|J|.$$
On the other hand, for any $y,y'\in J$ we have
$f(y)-f(y')< |f'(x_0)|(y-y')+\varepsilon(y-x_0)+\varepsilon(y'-x_0) \le |f'(x_0)||J|+2\varepsilon|J|$.
This implies
$$|f(J)|\le |f'(x_0)||J|+2\varepsilon|J|.$$
Together we have
$$\left|\frac{|f(J)|}{|J|}-|f'(x_0)|\right| < 2\varepsilon$$
for any interval $J$ which has endpoint $x_0$ and small enough length.