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I know that $\sqrt{|xy|}$ is continuous function but i'm not getting how to prove this. Do i have to use polar coordinates or to select any path like y=mx to check the continuity??

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    MathJax hint: if you put braces around the stuff that goes under the square root sign, the top bar expands to cover it. Check my edit. – 2017-01-24
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    You just have to prove $xy$ is continuous. The rest follows by composition. – 2017-01-24
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    Check out this question: http://math.stackexchange.com/questions/339289/continuous-function-proof-by-definition – 2017-01-24
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    Prove that if f and g are continuous real functions then h(x,y)=f(x)g(y) is continuous using standard epsilon-delta methods. – 2017-01-24

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As you say: let $x=r\cos\theta$ and $y=r\sin\theta$, where $r>0$.

\begin{eqnarray*} \sqrt{|xy|} &=& \sqrt{|r^2\cdot \cos\theta\sin\theta|}\\ \\ &=& \sqrt{\left|r^2\cdot \frac{\sin 2\theta}{2}\right|}\\ \\ &=& \frac{r}{\sqrt 2}\sqrt{|\sin 2\theta|} \end{eqnarray*}

Since $0\le |\sin 2 \theta| \le 1$ for all $\theta$, it follows that as $r \to 0$, $\sqrt{|xy|} \to 0$ for all $\theta$.

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By the inequality of arithmetic-geometric means $$\sqrt{|xy|}\leq\frac{|x|+|y|}{2}.$$ Now use the fact that $(x,y)\to(0,0)$ if and only if $|x|+|y|\to0$ to conclude that $\sqrt{|xy|}$ is continuous at $(0,0)$.