For $\epsilon > 0$, there exists $N\in \mathbb{N}$ such that $|y_n - y| < \epsilon$ for $n\geq N$. Removing the absolute value, adding $y$, then "inverting" the inequalities gives us $(y + \epsilon)^{-1} < y_n^{-1} < (y - \epsilon)^{-1}$.
We want $(y - \epsilon)^{-1} = \epsilon_0 + y^{-1}$ for $\epsilon_0 > 0$. If we solve for $\epsilon$ , we get $\epsilon = y(1 - (\epsilon_0y + 1)^{-1})$. I observed that, given a small enough $\epsilon_0>0$, $\epsilon$ is positive regardless of if $y$ is positive or negative. Then we can find a positive $\epsilon$, and a corresponding index $N_0 \in \mathbb{N}$, for an arbitrarily small $\epsilon_0$.
I would proceed by defining a third epsilon, $\epsilon^* = max\{|(y+ \epsilon)^{-1} - y^{-1}|,\epsilon_0\}$, observing that since the former expression tends to zero as $\epsilon_0$ tends to zero, $\epsilon^*$ can be made arbitrarily small. Then $|y_n^{-1} - y^{-1}|<\epsilon^*$ for $n\geq N_0$.
This is wrong. You cannot define a $\varepsilon^*$. You need to work with any $\varepsilon>0$.
First of all, note that there is a $N_1\in\mathbb N$ such that$$n\geqslant N_1\implies\lvert y-y_n\rvert<\frac{\lvert y\rvert}2\implies\lvert y_n\rvert>\frac{\lvert y\rvert}2.$$ But then, if $n\geqslant N_1$,$$\left\lvert\frac1y-\frac1{y_n}\right\rvert=\frac{\lvert y-y_n\rvert}{\lvert y\rvert\rvert y_n\rvert}\leqslant\frac2{\lvert y\rvert^2}\lvert y-y_n\rvert.$$Now, take $N_2\in\mathbb N$ such that $n\geqslant N_2\implies\lvert y-y_n\rvert<\dfrac{\varepsilon\lvert y\rvert^2}2.$ Then$$n\geqslant\max\{N_1,N_2\}\implies\left\lvert\frac1y-\frac1{y_n}\right\rvert<\varepsilon.$$