calculus-point $(0,0)$is continuous of $f\left(x,y\right)=\begin{cases}\sqrt{\left|xy\right|}\sin\left(\frac{1}{xy}\right)&xy\ne 0\\0&xy=0\end{cases}$

Question

$$f\left(x,y\right)=\begin{cases}\sqrt{\left|xy\right|}\sin\left(\frac{1}{xy}\right)&xy\ne 0\\ 0&xy=0\end{cases}$$
First of all, the question:
If the partial derives at $(0,0)$ exist, can I say automatically that it is continuous at $(0,0)$? I thought of proving that way since the question before was to show the partial derives existed at that point.
If not, then I tried proving it, can you say please if it is good?
For all $\epsilon>0$, there exist $\delta>0$, such that: $0<||(x,y)-(0,0)||<\delta$ which concoludes: $|f(x,y)-f(0,0)|<\epsilon$
So from $\delta$ I know: $||x||<\delta$ and that way with $y$ also.
now:
$|f(x,y)-f(0,0)| = |f(x,y)| = \left|\sqrt{xy}\cdot \sin\left(\frac{1}{xy}\right)\right| \le |\delta \cdot \sin\left(\frac{1}{xy}\right)| \le \delta $ and thus if we choose $\delta = \epsilon$, we will receive truth.
I have another solution which the professor said, but that is my try.

We know $||(x,y)|| < \delta$, thus:
$||x|| < \delta$, so with $\sqrt {||x||}$, it is $\sqrt \delta$
$||y|| < \delta$
or basically ( but have not used ): $x^2+y^2 < \delta$

Answer 1

I apply the $\delta$-$\varepsilon$ technique when I am really desperate, or the question requires using that method.

The function is continuous at each point $(x_0,y_0)$ such that $x_0y_0\neq 0,$ as it is represented as a composition, multiplication and division of continuous functions.

Concerning $x_0y_0=0$ we have $$|f(x,y)-f(x_0,y_0)|=|f(x,y)|=\begin{cases} 0 & xy=0\\ \sqrt{|xy|}\left |\sin{1\over xy}\right | & xy\neq 0 \end{cases}\ \le \sqrt{|xy|}$$ When $(x,y)\to (x_0,y_0),$ then $xy\to x_0y_0=0.$ Hence $\sqrt{|xy|}\to 0.$ This implies $|f(x,y)-f(x_0,y_0)|\to 0,$ when $(x,y)\to (x_0,y_0).$ In particular the case $x_0=y_0=0$ is covered.

Answer 2

Since $(|x|-|y|)^2\ge 0\iff |xy|\le\frac 12(|x|^2+|y|^2)\ $ we get

$$\sqrt{|xy|}\le \frac 1{\sqrt{2}}\sqrt{|x|^2+|y|^2}=\frac 1{\sqrt{2}}\sqrt{|x-0|^2+|y-0|^2}=\frac 1{\sqrt{2}}\Big\lVert(x,y)-(0,0)\Big\rVert_2$$

Also from $|\sin|\le 1$ we have $|f(x,y)|\le \sqrt{|xy|}$ it's quite straightforward to conclude.

I think this was probably the intent of the exercise to make you use the comparison between $\sqrt{|xy|}$ and the Euclidean norm.