Let $f : \mathbb{R}^2 \to \mathbb{R}$ be the function such that $$ f(x, y) = \begin{cases} \frac{(x^2 - y)y}{x^4}, & 0 < y < x^2\\ 0, & \text{elsewhere} \end{cases} $$ Is $f$ continuous?
My attempt:
I tried to use the characterizations of continuity by sequence. So I took the sequence $x_k = \frac{2}{k}$ and $y_k = \frac{1}{k^2}$ and so I had $(x_k, y_k) \rightarrow (0, 0)$ but $f(x_k, y_k) = \frac{3}{16}$ wich converges trivially to $\frac{3}{16}$. And this value is different from $f(0, 0) = 0$. So $f$ is not continuous at $(0,0)$ but what about the rest of the point? And also I don't know how to manipulate the condition $0 < y < x^2$ ir order to prove or disprove continuity.
Any hint or answer would be really appreciated.
We have that for $x\neq 0$ $f(x,y)$ is continuous that is
$$\lim_{y\to (x^2)^-}\frac{(x^2 - y)y}{x^4} =0 \quad \lim_{y\to 0^+}\frac{(x^2 - y)y}{x^4} =0$$
and for $(x,y)\to (0,0)$ by $x=t$ and $y=t^2-t$ with $t\to 0$
$$\frac{(x^2 - y)y}{x^4}=\frac{(t^2-t^2+t)(t^2-t)}{t^4}=\frac{t-1}{t^2}\to -\infty$$
therefore the function is continuous in $\mathbb{R^2}\setminus\{(0,0)\}$.