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?

Setting $$y=tx$$ then we get
$$\frac{x^3+t^3x^3}{x^2+t^2x^2}=x\frac{1+t^3}{1+t^2}$$ this tends to Zero if $x$ tends to zero
