$$f(x, y) = \begin{cases} \frac{x^4y^3}{x^8 + y^4} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0)\end{cases}$$
After having tried a couple of paths, getting $0$ as a limit, I took it as a good starting point, and thence I tried to prove the function is indeed continuous at the origin:
$$\bigg|\frac{x^4y^3}{x^8 + y^4} - 0 \bigg| = \frac{x^4y^2|y|}{x^8+y^4}$$
Now by AM/GM: $x^8 + y^4 \geq 2\sqrt{x^8y^4} = 2x^4y^2$ hence
$$\frac{x^4y^2|y|}{x^8+y^4} \leq \frac{x^4y^2|y|}{2x^4y^2} = \frac{1}{2}|y| \leq \frac{1}{2}(|y| + |x|) \leq C\frac{1}{2}\sqrt{x^2+y^2} \to 0$$
When $(x, y) \to (0, 0)$.
For some constant $C$ we do not care about (it's finite and positive). That shall be the norm in $\ell^2$: I studied that I can always perform this: $$|x| + |y| \leq C \sqrt{x^2+y^2}$$
So, did I well?
The last inequality proposed does not hold since \begin{align*} (|x| + |y|)^{2} = x^{2} + 2|xy| + y^{2} \geq x^{2} + y^{2} \geq 0 & \Rightarrow |x| + |y| \geq \sqrt{x^{2} + y^{2}} \end{align*}
In fact, it suffices to apply the squeeze theorem to the relation: \begin{align*} 0\leq \left|\frac{x^{4}y^{3}}{x^{8} + y^{4}}\right| \leq \frac{|y|}{2} \leq |y| \end{align*}
Hopefully this helps!