Given
$$f ({x,y})= \begin{cases} {{x^2y} \over {(x^2+y^2)^\alpha}},&(x,y) \ne {(0,0)}\\ 0,&(x,y)={(0,0)} \end{cases}$$
For what values of $\alpha$, $f$ is continuous in ${(0,0)}$?
I set $\space x=r \cos \theta$, $\space y=r \sin \theta$ and get $${{x^2y} \over {(x^2+y^2)^\alpha}} = r^{3-2 \alpha}\cos ^2 \theta\sin \theta $$
so it goes to $f(0,0)=0$ if $\space {3-2 \alpha}>0$, namely $\alpha < {3 \over 2}$.
But Can I say this is iff? And also the limit is not $f(0,0)=0$ (or not exist) if $\alpha \ge {3 \over 2}$?
It is true that the function is continuous at $(0,0)$ if and only if $\alpha\lt \frac{3}{2}$. However, saying "iff" is at best incomplete, one should discuss the cases $\alpha=\frac{3}{2}$ and $\alpha \gt \frac{3}{2}$ to indicate why the function is not continuous in these cases.
If $\alpha=\frac{3}{2}$, then our function is $\cos^2\theta\sin\theta$. If we approach $(0,0)$ along the path $\theta=\frac{\pi}{4}$, for example, then the limit as we approach $(0,0)$ along that path is $\frac{1}{2\sqrt{2}}$. If we approach $(0,0)$ along the path $\theta=0$ (the positive $x$-axis), then the limit along that path is $0$. So the behaviour of $f(x,y)$ near $(0,0)$ is path-dependent. In particular, the limit is not $0$, indeed does not exist.
If $\alpha \gt \frac{3}{2}$, we run into similar difficulties. If we approach $(0,0)$ along the path $\theta=\frac{\pi}{4}$, our function blows up as $(x,y)\to (0,0)$. If we approach along the path $\theta=0$, the function has limit $0$. So again the behaviour near $(0,0)$ is path-dependent.