Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix} 
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\ 
0 & , (x,y)=(0,0) 
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$. 
I have done the following:  
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$ 
Is this correct? 

Continue by observing that for any $\varepsilon > 0$, all values in the origin-less open disc of radius $\delta = \varepsilon/(2\sqrt{2})$ have absolute value $< \varepsilon$; this, in conjunction with $f(0, 0) = 0$, establishes the continuity of $f$ at $(0, 0)$.
