Conditions to exploit Polar coordinates in limits. 
Evaluate, $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^2y}{x^4+y^2}$$

When I used polar coordinates with $x=r\cos\theta, y=r\sin\theta$,
$$\lim_{r\rightarrow0}\dfrac{r\cos\theta\sin2\theta}{r^2\cos^4\theta+\sin^2\theta}=0$$
But when I use path $y=x^2$,
$$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^4}{2x^4}=1$$
Also from path $x=0$ or $y=0$ both gives,
$$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^2y}{x^4+y^2}=0$$
From path knowledge, I can say Limit does not exist.
Why this occurred that I got two different values of limits from Polar and the path makes me put a question that when to employ polar coordinates method to compute limits? When can I ascertain that it gives the correct value? Why is it giving out the value $0$ even when limit DNE?
Please help!

If in polar coordinates the function takes the form
$$
g(r) \, h(r,\theta)
$$
where $g(r) \to 0$ as $r \to 0^+$ (standard single-variable limit) and the function $h$ is bounded for all $\theta$ and all $r$ in some region $0 < r < R$, then you can draw the conclusion that the two-variable limit is zero.
But that's not what you have in your case. Sure, you get a factor $r$ which tends to zero, but the remaining expression isn't bounded (as your other argument with $y=x^2$ shows; no matter how small $r$ is, you can find a $\theta$ such that the whole expression equals $1$, i.e., that other part standing together with $r$ is equal to $1/r$, which is unbounded as $r \to 0$).
