Is this function continuous at $(0,0)$. The function is
$$
f(x, y) = 
\cases{        
\frac{2x^2y}{x^4+y^2} & (x, y) $\not=$ (0, 0)\cr
0 & (x, y)=(0, 0)\cr
}
$$
I want to know if this is continuous on $(0,0)$ or not. To do it, I can try to prove the partial derivatives $f_x, f_y$ are continuous on $(0, 0).$ Or I can prove this function is differentiable at $(0, 0).$
I used the second method and I need $$\lim\limits_{(x, y) \rightarrow(0, 0)}\frac{\frac{2x^2y}{x^4+y^2}}{\sqrt{x^2-y^2}}=0$$ to prove that it is differentiable.
So now my question is how I can prove this to be true.
 A: Well you need continuity first to be allowed to speak about derivatives...
In this type of exercise, you want to make the denominator homogeneous to be able to factor something.
You have $x^4$ and $y^2$ so let set $y=ux^2$
Note: $u$ is a variable not a constant, the ratio $\frac y{x^2}$ if you prefer.
$\require{cancel}f(x,y)=\dfrac{2x^2y}{x^4+y^2}=\dfrac{2\cancel{x^2}u\cancel{x^2}}{\cancel{x^4}+u^2\cancel{x^4}}=\dfrac{2u}{1+u^2}$
You can notice that this is not unconditionality converging to zero, for instance if $u=1$ the value is $f(x,x^2)=1\neq f(0,0)$ so the function is not continuous in $(0,0)$.
This method helps you finding paths along which the limit takes different values, instead of guessing which might work or not.

Remark if the exercise would be $f(x,y)=\dfrac{2x^3y}{x^4+y^2}$ instead, the same method gives you:
$f(x,y)=\underbrace{\dfrac{2u}{1+u^2}}_\text{bounded}\times \underbrace{x}_{\to 0}\to 0$ and this would be continuous.
A: The function is NOT continuous at origin.
Along the parabola $y=x^2$ is the limit of $\frac{2x^2y}{x^4+y^2}$ equal to $1,$
while along the coordinate axes it is $0.$
A: Other people have provided answers already but I think its also good to get some intuition of what such a function looks like and why it's not continuous. You can see that depending on how your approach the origin, you'll get different limits. In particular, there's some funky stuff going on at $y = kx^2$.

A: $\displaystyle\lim_{x\to 0} f(x,x^2)\ = 1,\ $ whereas $\displaystyle\lim_{x\to 0} f(x,2x^2)\ = \frac{4}{5},\ $ so no, $f$ is not continuous at $(0,0).$
