<s>[INST] Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists. 
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
  exists.

Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$ 
is this method correct?
 [/INST] You want to prove that for every $\epsilon>0$ there exists $\delta>0$ such that
$$0<\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^3}{x^2+y^2}\right|<\epsilon.$$
Let $\delta=\frac{\epsilon}{2}$. Then
$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{y^3}{x^2+y^2}\right|$$
$$=|x|\underbrace{\left|\frac{x^2}{x^2+y^2}\right|}_{\leq 1}+|y|\underbrace{\left|\frac{y^2}{x^2+y^2}\right|}_{\leq 1}$$
$$\leq|x|+|y|\leq\sqrt{x^2+y^2}+\sqrt{x^2+y^2}$$
$$=2\sqrt{x^2+y^2}<2\cdot\frac{\epsilon}{2}=\epsilon$$
This proves that the limit exists and is equal to $0$.
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