Finding a sequence to show discontinuity of $f(x,y)$

Question

Let $f: \mathbb R^{2} \to \mathbb R$, with $f(x,y) = \begin{cases} xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}} & (x,y)\neq 0\\ 0 & (x,y)=0 \\ \end{cases}$

I have already proved that $f$ is 2-times partially differentiable and subsequently $\partial_{1}\partial_{2}f(0,0)\neq\partial_{2}\partial_{1}f(0,0)$.

Next, I am asked to prove or refute that $f$ is continuous in $(0,0)$.

From the fact $\partial_{1}\partial_{2}f(0,0)\neq\partial_{2}\partial_{1}f(0,0)$ I assume that it is not continuous in $(0,0)$, so I am attempting to find a sequence like $(\frac{1}{n},\frac{1}{n^{2}})$ such that either $\lim_{n\to \infty}f(\frac{1}{n},\frac{1}{n^{2}})\neq f(0,0)$ but every null sequence I have tried renders $\lim_{h \to 0}f(h,h)=f(0,0)$, which leads me to believe that it may be continuous?

Furthermore if I can prove that $f$ is discontinuous in $f(0,0)$, am I correct in saying that $\nabla f$ cannot be continuous in $(0,0)$, if not how can I prove it?

Answer 1

$$ \left|f\left(x,y\right)-0 \right| \leq \left|xy\right| \leq \left\|\left(x,y\right)\right\|^2 $$

So $f$ is continuous at $(0,0)$ However, with Schwarz theorem, you know $f$ cannot be $\mathscr{C}^2$ on $\left(\mathbb{R}^{*+}\right)^2$.

Answer 2

It turns out that $f$ is continuous at $(0,0)$. In fact,$$0\leqslant\left|xy\frac{x^2-y^2}{x^2+y^2}\right|\leqslant|x|.|y|\frac{x^2+y^2}{x^2+y^2}=|x|.|y|.$$Therefore, the continuity of $f$ at $(0,0)$ follows from the squeeze theorem.