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...
 A: Note that we have
$$\frac12|2x+3y|\le |x|+\frac32|y|\le \frac52\sqrt{x^2+y^2}<0.01$$
whenever $\sqrt{x^2+y^2}<\delta=0.004$
A: To get a larger possible value of $\delta$, use the polar coordinate transformation $x = r \cos t, y = r \sin t$ to rewrite the equality in the question.
$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{r^3 \cos t \sin t (2\cos t+3\sin t)}{r^2}\right\vert = r |\cos t| |\sin t| |2 \cos t + 3 \sin t|
$$
In another answer, choosing $|x|,|y| \le \sqrt{x^2+y^2}$ is similar to bounding $|\sin t|,|\cos t|$ by $1$.  IMHO, it's too brutal as all information about the variable $t$ is lost.  $2 \cos t + 3\sin t = \sqrt{2^2+3^2} \sin(t + \alpha) = \sqrt{13} \sin(t+\alpha)$, where $\tan \alpha = \dfrac23$.
From this
$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}\right\vert
= \frac{\sqrt{13}}{2} r |\sin 2t \sin(t+\alpha)| \le \frac{\sqrt{13}}{2} \delta < \epsilon$$
for any $t \in [0,2\pi]$ and $r < \delta$.
The choice $\delta < \dfrac{2}{\sqrt{13}} \epsilon$ guarantees that $|f(x,y)-f(0,0)| < \epsilon$.  When $\epsilon = 0.01$, it becomes
$$\delta < \frac{2}{\sqrt{13}} \cdot \frac{1}{100} = \frac{1}{50\sqrt{13}} \approx 0.005547. \quad \text{(cor. to 4 d.p.)}$$
This improves the bound given in another answer.

The graph of $g(t) = \cos t \sin t (2 \cos t + 3 \sin t)$

Observe that the amplitude is between $1.5$ and $2$.  This suggest that choosing $\dfrac52 \delta < \epsilon$ is too restrictive.  We can allow more points to enter this $\delta$-ball in the domain of $f$ by choosing a larger $\delta$.  For a fixed $\epsilon$, this means choosing a smaller coefficient of $\delta$.  The choice $\dfrac{\sqrt{13}}{2} \approx 1.8028$ is nearer to the amplitude than $2.5$.

Another answer shows that (2) "$\delta < 0.001$" is true and (4) "no such $\delta$ exists" is false.  To complete this question, it remains to discuss the possibility of (1) and (3).  To illustrate the sharpness of the bound $\delta < \dfrac{1}{50 \sqrt{13}}$ given above, choose $(r,t) = \left(0.0057, \dfrac{13}{50} \pi\right)$.  This gives $(x,y) \approx (0.0039, 0.0042)$, and $f(x,y) \approx 0.010114 > 0.01 = \epsilon$.
Hence, (1) and (3) do not hold.
