How to prove that $\lim\limits_{(x,y) \to (0,0)} \frac{\left | x \right |^{\frac{3}{2}}y^{2}}{x^{4} + y^{2}} \rightarrow 0$

I assume you want to show that

$$\lim_{(x,y)\rightarrow (0,0)}\frac{|x|^\frac{3}{2}y^2}{x^4+y^2}= 0.$$ Although sometimes it is not the nicest way, switching to polar coordinates is a general approach that will solve these kinds of limits. (and we don't have to think too much!) Let $x=r\cos \theta$, $y=r\sin\theta$. Then your limit is $$\lim_{(x,y)\rightarrow(0,0)}\frac{|x|^{\frac{3}{2}}y^{2}}{x^{4}+y^{2}}=\lim_{r\rightarrow0}\frac{r^{\frac{3}{2}}|\cos\theta|^{\frac{3}{2}}r^{2}\sin^{2}\theta}{r^{2}\left(r^{2}\cos^{4}\theta+\sin^{2}\theta\right)}=\lim_{r\rightarrow0}r^{\frac{3}{2}}\frac{|\cos\theta|^{\frac{3}{2}}}{\left(r^{2}\cos^{4}\theta/\sin^{2}\theta+1\right)}.$$

Now, since

$$0\leq \frac{|\cos\theta|^{\frac{3}{2}}}{\left(r^{2}\cos^{4}\theta/\sin^{2}\theta+1\right)}\leq |\cos\theta|^\frac{3}{2}\leq 1$$

we can apply the squeeze theorem, and since $\lim_{r\rightarrow 0}r^\frac{3}{2}=0$, we see that the original limit is $0$.

Hope that helps,

For $y \neq 0$, $$ 0 \le \frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} = \frac{{|x|^{3/2} }}{{x^4 /y^2 + 1}} \le \frac{{|x|^{3/2} }}{1} = |x|^{3/2} \to 0. $$

EDIT: In retrospect, simply note that $0 \le \frac{{y^2 }}{{x^4 + y^2 }} \le 1$, for $(x,y) \neq (0,0)$, to conclude that $\frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} \to 0$ as $x \to 0$, independently of $y$.