1. **State the problem:** Given $\epsilon = 2023$, find $\delta > 0$ such that for all $(x,y)$, we have $$|f(x,y) - f(0,0)| < \epsilon$$ whenever $$x^2 + y^2 < \delta,$$ where $$f(x,y) = \frac{y}{x^2 + 1}.$$ 2. **Evaluate $f(0,0)$:** $$f(0,0) = \frac{0}{0^2 + 1} = 0.$$ 3. **Rewrite the inequality:** We want $$\left| \frac{y}{x^2 + 1} - 0 \right| = \left| \frac{y}{x^2 + 1} \right| < 2023$$ whenever $$x^2 + y^2 < \delta.$$ 4. **Analyze the function:** Note that for all real $x$, $$x^2 + 1 \geq 1,$$ so $$\left| \frac{y}{x^2 + 1} \right| \leq |y|.$$ 5. **Use the condition on $(x,y)$:** Since $$x^2 + y^2 < \delta,$$ we have $$|y| \leq \sqrt{x^2 + y^2} < \sqrt{\delta}.$$ 6. **Combine inequalities:** Therefore, $$\left| \frac{y}{x^2 + 1} \right| < \sqrt{\delta}.$$ 7. **Find $\delta$ to satisfy the inequality:** We want $$\sqrt{\delta} < 2023,$$ which implies $$\delta < 2023^2 = 4092529.$$ 8. **Conclusion:** Any $$\delta > 0$$ such that $$\delta < 4092529$$ will satisfy the condition $$|f(x,y) - f(0,0)| < 2023$$ whenever $$x^2 + y^2 < \delta.$$