Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined as $$(x,y) \longmapsto \left\{ \begin{array}{cl} \dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & \mbox {if } (x,y) \neq (0,0) \\ \\ 0 & \mbox {if } (x,y) = (0,0) \end{array} \right. $$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $\epsilon-\delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $\epsilon > 0$ I need to find a $\delta$ such that $\|(x,y)\| =\sqrt{x^2 + y^2} < \delta$ implies $|f(x,y)| < \epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 \geq 4x^2y^2$ $$\begin{align*} \left|\frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}\right|&=\left|\frac{4x^2y^3}{(x^2+y^2)^2} + \frac{x^4 y}{(x^2+y^2)^2} - \frac{y^5}{(x^2+y^2)^2}\right|\\ &\leq\left|\frac{4x^2y^3}{(x^2+y^2)^2}\right| + \left|\frac{x^4 y}{(x^2+y^2)^2}\right| + \left|\frac{y^5}{(x^2+y^2)^2}\right|\\ &\leq\left|\frac{4x^2y^3}{4x^2y^2}\right| + \left|\frac{x^4 y}{4x^2y^2}\right| + \left|\frac{y^5}{4x^2y^2}\right|\\ &= \left|y\right| + \left|\frac{x^2}{4y}\right| + \left|\frac{y^3}{4x^2}\right| \end{align*}$$
Got stuck here and not sure if my approach is correct.
