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?

Since
\begin{align}
\left|\frac{x^3+y^3}{x^2+y^2}\right|&=\left|(x+y)\cdot\frac{x^2-xy+y^2}{x^2+y^2}\right|\\
&=|x+y|\left|1-\frac{xy}{x^2+y^2}\right|\\
&\le|x+y|\left(1+\frac{|xy|}{x^2+y^2}\right)\quad(\because\text{the triangle inequality})\\
&\le|x+y|\left(1+\frac{1}{2}\right)\quad(\because\text{the AM-GM inequality})\\
&=\frac{3}{2}|x+y|
\end{align}
and $\displaystyle \lim_{(x,y)\to(0,0)}\frac{3(x+y)}{2}=0$, $\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^3+y^3}{x^2+y^2}=0$ by the Squeeze Theorem.
