$F(x,y)$ is continuous. 
Prove that
  $$ f(x,y)=\begin{cases}\frac{x^3-xy^2}{x^2+y^2}&\text{if }(x,y)\ne(0,0)\\0&\text{if }(x,y)=(0,0)\end{cases}$$
  is continuous on $\mathbb R^2$ and has first partial derivatives everywhere on $\mathbb R^2$, but is not differentiable at $(0,0)$.

I want to show that the function is continuous on $\Bbb R^2$ 
For that, It is sufficient to show that the function is contiously differentiable on $\Bbb R^2$ 
Does there exist another choice to prove this? 

Edit I added second question. 
Use differentials to approximate $(597)(16.03)^{1/4}$ 

There, how to find the value of $a$ , $b$ , $dx $ and $dy$ 
Please explain me thanks :) 
 A: Hint: To show  that $f$ is continuous at $(0,0)$, use polar coordinates ie. $\left\{ \begin{array}{l} x=r \cos(\theta) \\ y=r \sin(\theta) \end{array} \right.$.
A: Since $|x|,|y|\leq \sqrt{x^2+y^2}=||(x,y)||$ then
$$\frac{|x^3-xy^2|}{x^2+y^2}\leq \frac{|x|^3+|x|y^2}{x^2+y^2}\leq \frac{||(x,y)||^3+||(x,y)||||(x,y)||^2}{||(x,y)||^2}\leq2||(x,y)||\to0=f(0,0)$$
so $f$ is continuous at $(0,0)$ and hence everywhere.
A: Proving that it is continuously differentiable is WAY overkill - and in this case impossible, given that the problem directly asks you to show that it isn't differentiable.  Just prove continuity directly!
It is immediate that this function is continuous at all points other than $(0,0)$, since rational functions are continuous on their domain.  So, all that you need to show is that
$$
\lim_{(x,y)\rightarrow(0,0)}f(x,y)=f(0,0).
$$

Edit: A helpful trick for showing that the function is continuous at $(0,0)$: note that since $f(0,0)=0$, it is enough to show that $\lvert f(x,y)\rvert$ tends to 0 as $(x,y)\rightarrow(0,0)$. Now, we have that for $(x,y)\neq(0,0)$, 
$$
\newcommand{\abs}[1]{\left\lvert #1\right\rvert}\abs{f(x,y)}=\abs{\frac{x^3-xy^2}{x^2+y^2}}=\abs{\frac{x(x-y)(x+y)}{x^2+y^2}}.
$$
But, we have that $\abs{x}=\sqrt{x^2}\leq\sqrt{x^2+y^2}$; similarly,
$$
\abs{x+y}\leq\abs{x}+\abs{y}\leq\sqrt{x^2+y^2}+\sqrt{x^2+y^2}=2\sqrt{x^2+y^2}.
$$
So, we find that for $(x,y)\neq(0,0)$,
$$
\abs{f(x,y)}\leq\frac{2(x^2+y^2)\abs{x-y}}{x^2+y^2}=2\abs{x-y}\rightarrow0\text{ as }(x,y)\rightarrow(0,0).
$$
Depending on what your instructor wants from you, this might be overkill - in that case, I'd suggest either going with Seirios' suggestion or talking to your instructor about what they want to see.
