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.