What's wrong with the method I used to prove this limit doesn't exist? The limit is 

$\lim_{x\to0,y\to0}\frac{x^2y}{x^2+y^2}$

Here is the way to prove the limit is $0$:

For arbitrary $\epsilon$, there exists a $\delta $ ($ 0 &lt; \sqrt{x^2+y^2} &lt; \delta$) that can satisfy the equation $| {\frac {x^2y} {x^2+y^2}}|&lt;\epsilon$ as $x^2 \le x^2+y^2$ hence $|\frac{x^2y}{x^2+y^y}| \le |y| = y^2\le \sqrt {x^2+y^2}$ if we choose $\delta = \epsilon$.

Nevertheless, I try to choose (x, y) approaching (0,0) along a particular curve such as $x^2=-y^2+y$ and $x^2=-y^2+y^3$, which give me the limits are $0$ and $-1$ respectively that can prove the limit doesn't exist.

The question I wanna ask is what's wrong the curve I chosen and what is the rule to or not to choose a particular curve. 

Any help will be appreciated, thank you!
You can do this also:

Put $$x=r\cos{t}$$ $$y=r\sin{t}$$

Then:

$$\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r \to 0} \frac{r^3 \cos^2{t}\sin{t}}{r^2}=\lim_{r \to 0}r \cos^2{t}\sin{t} = 0$$,
$\forall t \in [0,2\pi)$

Thus the limit is zero.

In general it is a very useful technique to use polar coordinates for computing multivariable limits or to examine the existence of a limit.

For instanse if the result  as $r \to 0$ depends on values of $t$ then the limit does not exist.

As an exercise for you try to prove that $\lim _{(x,y) \to (0,0)} \frac{x-y}{x+y}$ does not exist,using polar coordinates.
You can do this also:

Put $$x=r\cos{t}$$ $$y=r\sin{t}$$

Then:

$$\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r \to 0} \frac{r^3 \cos^2{t}\sin{t}}{r^2}=\lim_{r \to 0}r \cos^2{t}\sin{t} = 0$$,
$\forall t \in [0,2\pi)$

Thus the limit is zero.

In general it is a very useful technique to use polar coordinates for computing multivariable limits or to examine the existence of a limit.

For instanse if the result  as $r \to 0$ depends on values of $t$ then the limit does not exist.

As an exercise for you try to prove that $\lim _{(x,y) \to (0,0)} \frac{x-y}{x+y}$ does not exist,using polar coordinates.