Hi Julio,
To prove the result, assume the hypothesis and manipulate the expression $|\frac1{y} - \frac1{y_0}|$ to a get a term involving $|y - y_0|$ as follows:
$(*) \displaystyle \left|\frac1{y} - \frac1{y_0}\right| = \left|\frac{y_0 - y}{yy_0}\right| = \frac{|y - y_0|}{|yy_0|}$.
These steps will now be justified. Since
$\displaystyle |y - y_0| < \min\left(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{2}\right)$,
in particular
$\displaystyle |y - y_0| < \frac{|y_0|}{2}$.
This inequality implies $y \neq 0$. Otherwise, $|y_0| < \frac{|y_0|}{2}$, which is contradiction. Thus $yy_0 \neq 0$ and the equations in $(*)$ are valid. Again by assumption,
$\displaystyle |y - y_0| < \frac{\varepsilon |y_0|^2}{2}$.
Further, by the triangle inequality, $|y - y_0| < \frac{|y_0|}{2}$ implies $|y| \ge |y_0| - |y - y_0| > \frac{|y_0|}{2}$. Hence
$\displaystyle \frac{|y - y_0|}{|yy_0|} < \frac{\varepsilon |y_0|^2}{2} \frac{2}{|y_0|^2} = \varepsilon$.