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I used polar coordinates to solve the problem and reached up to

$$ r|\cos^3(\theta) + 2\sin^3(\theta)| $$

I am stuck after that and don't know how to figure out epsilon and delta.

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    Are you required to show it using epsilons and deltas, or may you resort to the fact that $f$ is continuous at $(a,b)$ iff $lim_{(x,y)\rightarrow (a,b)} f(x,y) = f(a,b)$ in all directions? – 2017-01-31
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    $\cos^3(\theta)+2\sin^3(\theta)$ is a bounded function, hence $f\to 0$ as $(x,y)\to(0,0)$. – 2017-01-31

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You did not give the value $f(0,0)$. Nevertheless, we have the product of two factors. One of them ($r$) tends to $0$, while the second one (trigonometric part) is bounded independently on $\theta$. So, the whole product tends to $0$ and defining $f(0,0)=0$ we arrive at the continuous function.

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Hint: for any $\theta$,

$$0\le r|\cos^3(\theta)+2\sin^3(\theta)|\le3r$$

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    Thanks I got that. So after that we get 3sqrt(x^2+y^2) < epsilon i.e., |x| < sqrt( epsilon/6 ) and |y| < sqrt( epsilon/6). Am I going in right direction ? – 2017-01-31
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    @Rahul yup, looks good! – 2017-01-31
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    With a bit of work we could even get a tighter upper bound of $2r$, since the bounded part has a maximum at $2$. Not that it matters. – 2017-01-31