I need to investigate at which points the following function is continuous:
It is clear that for any point with $y \neq x^{2} $, it is continuous. I suspect that along the curve $y =x^{2} $ it is not continuous. For specific points, I used to prove discontinuity by sequential characterization, for instance: considering the sequence $(1, \frac{n}{n+1} )$ which converges to $(1,1)$, it is easy to show that f is not continuous at $(1,1)$. But how to generalize this argument?
Using @QC-QAOA 's idea, I wrote down a proof but as I am not sure any feedback is appreciated: $ \forall x \neq 0$ take the sequence $ (x, x^{2}+ \frac{1}{n} ) \rightarrow (x, x^{2} )$. Then $ f( x_{n} , y_{n} ) = y_{n} = (x^{2}+ \frac{1}{n}) \rightarrow x^{2} \neq 0 = f(x, x^{2} )$. By sequential characterization of continuity, f is discontinuous at all points $(x, x^{2} )$ with $x \neq 0$.
However, f is continuous at $(0,0)$: Take any sequence $( x_{n} , y_{n} ) \rightarrow (0,0)$. Then $(y_n) \rightarrow 0 $. Notice that $f( x_{n} , y_{n} ) \leq | y_{n} | \rightarrow 0 $. So, $f( x_{n} , y_{n} ) \rightarrow 0 $