I'll try to be more elaborate on the point I was trying to get across on the IRC earlier today.
Let's consider the following function, defined in the polar coordinates: $f: [0, \infty) \times (0, 2\pi] \to \mathbb{R}$,
$$
f(r, \theta) = \frac{r}{\theta}
$$
This is well defined, because our angle argument, $\theta$, varies in $(0, 2\pi]$, and not in $[0, 2\pi)$ as usual. Obviously this does not matter at all, because $2\pi$ determines the same angle as $0$. This function determines a function $\mathbb{R}^2 \to \mathbb{R}$ in a natural way, and I will denote that function by $f$ as well, using $x, y$ coordinates instead of $r, \theta$, hoping that will not cause any confusion. Also, $f(0) = 0$ (because the point $(0, 0) \in \mathbb{R}^2$ is represented in polar coordinates as $(0, \theta)$ for any choice of $\theta$, and $f(0, \theta) = \frac{0}{\theta} = 0$).
So, what's the limit:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \;?
$$
Let's try to solve it using your approach:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{r \to 0} \frac{r}{\theta} = 0
$$
Easy, wasn't it? And the resulting value does not depend on $\theta$. Let's conclude thus that the limit is $0$ and call it a day.
Right? Wrong! Consider a sequence (in polar coordinates):
$$
x_n = \left(r_n, \theta_n\right) = \left(\frac{1}{n}, \frac{\pi}{n}\right)
$$
Basically, we start with $x_1$ which is $(-1, 0)$ in cartesian coordinates, and we go clockwise, approaching half a turn, and going closer and closer to $(0, 0)$ in process. It's obvious that as a sequence of point in a plane, this sequence tends to $(0, 0)$ (because, well, in cartesian coordinates it's $x_n = (r_n \cos \theta_n, r_n \sin \theta_n) = (\frac{1}{n} \cos(\frac{\pi}{n}), \frac{1}{n} \sin(\frac{\pi}{n}))$, and obviously each component tends to $0$ as $n \to \infty$).
Now, if $\lim_{(x, y) \to (0, 0)} f(x, y)$ was in fact $0$, then we would also have $\lim f(x_n) = 0$, because $x_n \to (0, 0)$. But, surprisingly:
$$
\lim f(x_n) = \lim f(r_n, \theta_n) = \frac{\frac{1}{n}}{\frac{\pi}{n}} = \frac{1}{\pi} \ne 0
$$
What happened? It turns out, that $\lim_{r \to 0} f(r, \theta) = 0$ does not imply that $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$. When we consider the limit as $r \to 0$, we keep $\theta$ fixed, so we only take a limit along a single straight line passing through the origin. This only implies that if we restrict the our function to any such straight line $L$, and consider a function $f|_L: L \to \mathbb{R}$ it will in fact be continuous at $0$. The problem is that the fact that the restricted function $f|_L: L \to \mathbb{R}$ is continuous at origin for every $L$ passing through origin, it's not enough to conclude that our function is in fact continuous at origin.
Best Answer
The function is continuous in origin.
But you didn't demonstrate that the limit is equal to zero. Restricting it to just $y=kx^2$ is just one way you can approach the origin, you cannot say that the limit exists. The only conclusion from your result is that if the limit exists it must be $0$.
To demonstrate it formally you can use the Squeeze theorem as follows: $$\lim_{(x,y)\to(0,0)}\bigg|\frac{x^2\cdot |y|}{x^2+y^4}\bigg| \le \lim_{(x,y)\to(0,0)}\bigg|\frac{x^2\cdot |y|}{x^2}\bigg| $$ $$ \lim_{(x,y)\to(0,0)}\frac{{x^2}\cdot |y|}{x^2}=\lim_{(x,y)\to(0,0)}|y|=0 $$
We formally demonstrated that limit equals to $0$ when $(x,y)\rightarrow(0,0)$. As it is equal to the value of the function by definition of the function we can say that it is continuous.