Continuity of a piecewise function - considering paths

Question

Let the function $f$ be such that $$f(x,y) = \frac{x^2y}{x^4+y^2} \hspace{2mm} \text{,} \hspace{2mm} (x,y) \neq 0$$

and $0$ at the origin.

The question is: Is $f$ continuous at the origin?

The answer is no.

And it is proven in my notes by saying that $f$ vanishes along both coordinate axes and goes to zero along the straight line through the origin, but along the parabolic path $\gamma(t) = (t,t^2)$, $$f(\gamma(t)) = \frac{1}{2} \rightarrow \frac{1}{2} \neq 0 \hspace{2mm} \text{as} \hspace{1mm} t \rightarrow 0$$

However, why are we allowed to choose such paths? I mean, for example, $f(x) = x^2$ is continuous everywhere, and is 0 at the origin, but now let me consider the path $\gamma(t) = e^t$, then $f(\gamma(t)) = e^{2t}$ which goes to $1\neq 0$ as $t\rightarrow 0$?

Answer 1

When we choose the path, say, $y=x^2$, we are actually composing the function $f$ with the function $\gamma$ that you mentioned. And $\lim_{t\to0}f\bigl(\gamma(t)\bigr)=\frac12$. Since $\gamma$ is clearly continuous at $0$, we get from this that if $f$ is continuous at $(0,0)$, then $f(0,0)=\frac12$. But $f(0,0)$ is actually $0$. So, we can deduce that $f$ is not continuous at $(0,0)$.

Answer 2

Choosing a change of variables

$$ x = r\cos(\phi)\\ y = r^2\cos(\phi)\sin(\phi) $$

we get

$$f(x,y) = \frac{x^2y}{x^4+y^2} \hspace{2mm} \text{,} \hspace{2mm} (x,y) \neq 0 \equiv g(\phi) = \cos(\phi)\sin(\phi) $$

so the limit depends on the path taken and then does not exists.