$f(x,y)=x^y$ is not continuous at (0,0) How to prove that the function $f(x,y)=x^y$ ($x&gt;0$) is not continuous at the point $(0,0)$?

I tried $y=x^\alpha$ $(\alpha&gt;0)$ but this does not work since for such $y$ we have $x^y \to 1$ as $x\to 0$.

Thank you in advance!
Try 

$$y={-a\over\ln x}\qquad a\gt0$$

Then $y\gt0$ if $x\lt1$ and $y\to0$ as $x\to0$, but $x^y=e^{-a}$.
It is continous iff $$\lim_{(0,0)} y\log x$$ exists, but 

$$\lim_{x\to 0}\lim_{y\to 0} y\log x = 0\\
\lim_{y\to 0^+}\lim_{x\to 0} y\log x = -\infty \\
$$
Hence, the limit does not exist.