two variable function How can I show that this problem isn't differentiable at $(0,0)$?
at first, I think that this function isn't continuous at $(0,0)$ that implies it's not differentiable. Am I right?
Is there any other solution to prove whether this function is differentiable?
Thanks.
$$f(x,y) = \begin{cases} 2xy\frac{x^2-y^2}{x^2+y^2}, & \mbox{if } x^2+y^2 \neq 0 \\ 0, & \mbox{otherwise}\end{cases}$$
 A: The OP has a duplicate here Differentiability at $(0,0)$.
However, I think it could be useful make a general comment here on the topic of Differentiability and Continuity for functions of several variables. Any other contribution or observation will be appreciate.
1. Firstly you might check whether or not f is continuos in (0,0)
    by calculating the limit as $(x,y)\to(0,0)$. If it is discontinuos then
    it can't be differentiable, infact:
NOTE
continuity is a necessary condition since differentiability implies continuity
2. Then you have to check whether or not partial derivatives in (0,0)
        exist by calculating them as limits of the incremental ratio; if
        the partial derivatives do not exist then $f$ can't be
        differentiable, infact:
NOTE
existence of partial derivatives is a necessary condition since differentiability implies their existence
Even is $f$ is continuos in $(0,0$ and partial derivatives exist in $(0,0)$ you can't conclude anything yet about differentiability at $(0,0)$. 
3. To be sure you need to check that partial derivatives are continuos
    in $(0,0)$ by calculating $f_x$ and $f_y$ for $(x,y)\neq(0,0)$ and the
    calculationg their limit at (0,0). If their are continuos you are
    done, infact:
NOTE "Differentiability theorem"
if all the partial derivatives exist and are continuous in a neighborhood of (0,0) then (i.e. sufficient condition) the function is differentiable at $(0,0)$
4. If partial derivatives are not continuos at $(0,0)$ you can't yet
    conclude anything about differentiability. You need to check
    directly differentiability by definition that:
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$
NOTE
In this case you can also skip "3" and try directly with "2" and"4" to check differentiability.
A: From $f(x,0)=f(0,y)\equiv0$ it follows that $f_x(0,0)=f_y(0,0)=0$. This implies that $df(0,0)=0$ if $f$ is at all differentiable at $(0,0)$. We therefore have to check whether
$$\lim_{(x,y)\to(0,0)}{f(x,y)-f(0,0)-0\over\sqrt{x^2+y^2}}=0\ .$$
To this end we express $f$ in polar coordinates $r$, $\phi$ and obtain
$${f(x,y)-f(0,0)-0\over\sqrt{x^2+y^2}}={\hat f(r,\phi)\over r}={1\over2} r\sin(4\phi)\to0\qquad(r\to0+)\ .$$
This shows that in fact $df(0,0)=0$.
