<s>[INST] What is the value of $\delta $ if $\epsilon=0.01$? 
Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2},  & \text{if
 $(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$
Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0)
\vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-
1.$\delta <0.01$
2.$\delta <0.001$
3.$\delta <0.02$
4.no such $\delta $ exists

solution:since $f(0,0)=0$,then consider $\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{xy(2x+3y)}{x^2+y^2}\right\vert\le \frac{(2x+3y)}{2}$ as $xy\le \frac{x^2+y^2}{2}$
From here, how to proceed further...
 [/INST] Note that we have
$$\frac12|2x+3y|\le |x|+\frac32|y|\le \frac52\sqrt{x^2+y^2}<0.01$$
whenever $\sqrt{x^2+y^2}<\delta=0.004$
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