I am trying to prove $$\lim _{(x, y) \rightarrow(0,0)} \frac{3 x y^{2}}{x^{2}+y^{2}}=0$$ Using $\epsilon-\delta$ approach.
Given an $\epsilon >0$ we should find $\delta >0$ such that $$|f(x,y)-0|<\epsilon$$ Whenever $$0<\|(x, y)\|_{2}<\delta$$
Now i started as: $$\left|\frac{3 x y^{2}}{x^{2}+y^{2}}\right|<\varepsilon \to (1)$$ Also we have: $$\left|\frac{3 x y^{2}}{x^{2}+y^{2}}\right|<3|x|<3 \sqrt{x^{2}+y^{2}} \to (2)$$
But how to say from $(1)$ and $(2)$ $\delta=\frac{\epsilon}{3}$?
The trick is to note that $$0\leq \frac{y^2}{x^2+y^2} \leq 1,$$and that $|x| \leq \|(x,y)\|$. So, let $\epsilon > 0$. Pick $\delta = \epsilon/3$. If $0<\|(x,y)\| < \delta$, then in particular $|x|<\delta$, so $$\left|\frac{3xy^2}{x^2+y^2}\right| = 3|x|\frac{y^2}{x^2+y^2} \leq 3|x| < 3\delta = \epsilon.$$