<s>[INST] Analyze the continuity of the following function Here in my book I have such an exercise with the explanation given below, but still there is something the authors didn't add, but simply put "...after some operations...". Here is such an exercise:
Analyze the continuity of the following function:

$$f:\Bbb R^2\to\Bbb R $$    $$ f(x,y)= \begin{cases}
\frac{3x^2y}{\sqrt{x^4+y^2}}& \text{if } (x,y)\neq(0,0)\\[2ex]
0& \text{if }   (x,y)=(0,0)
\end{cases}$$
In my book is writen, after some operations, we find that  $f$ is continuous on $\Bbb R^2 \setminus\{(0,0)\}$. What are these operation ? Thanks.
 [/INST] if $|y|\leq 1$, $$\sqrt{x^4+y^2}\geq \sqrt{x^4+y^4}$$
and thus
$$|f(x,y)|=\left|\frac{3x^2y}{\sqrt{x^4+y^2}}\right|\leq\underbrace{\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|}_{\leq 1}|3y|\leq |3y|\underset{(x,y)\to(0,0)}{\longrightarrow 0}$$
therefore $f$ is continuous on $\mathbb R^2$.
Justification for $\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|\leq 1$: 
$$x^4\leq x^4+y^4\implies x^2\leq\sqrt{x^4+y^4}\implies \underbrace{\frac{x^2}{\sqrt{x^4+y^4}}}_{=\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|}\leq 1.$$
</s>