Answer
$|f(x,y) -f(0,0) | \lt \epsilon $
Work Step by Step
We have $f(x,y)=\dfrac{y}{x^2+1} ; \\f(0,0)=0$
Now, $|f(x,y) -f(0,0) | \lt \epsilon $ $ \implies |\dfrac{y}{x^2+1}-0| \lt 0.05 $
$\implies \dfrac{|y|}{x^2+1} \lt 0.05$
Since,$\dfrac{|y|}{x^2+1} \leq \sqrt {x^2+y^2 }$
Now, $\sqrt {x^2+y^2 } \lt 0.05$
Consider $\delta =0.05$ and $\sqrt {x^2+y^2 } \lt \delta$
Thus, we have $|f(x,y) -f(0,0) | \lt \epsilon $
