Differentiability at a point $(0,0)$ How would i show $$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}$$ is not differentiable at $(0,0)$
 A: I'm assuming you mean to set $f(0,0) = 0$. In that case, the function most definitely is continuous, so trying to show it's not won't work :)
If it were differentiable, the derivative would have to be given by the matrix
$$Df(0,0)=\begin{bmatrix}\frac{\partial f}{\partial x}(0,0) & \frac{\partial f}{\partial y}(0,0)\end{bmatrix}.$$
Both these partial derivatives are $0$ (note that $f(x,0) = f(0,y) = 0$ for all $(x,y)$). On the other hand, if we try to compute the directional derivative in the direction of the vector $\mathbf v=(2,1)$ we obtain
$$\lim_{t\to 0} \frac{f(2t,t)-f(0,0)}{t} = \lim_{t\to 0} \frac{6t^4}{t|t|^3},$$
which might look like $6$, but in fact does not exist.
Caveat: I do not know what sort of course you're taking. In some courses, directional derivatives require unit vectors; in other courses, they do not. Either way, it doesn't matter here.
But it's a basic Theorem that if $f$ is differentiable at $(0,0)$ we can compute the directional derivative at in direction $\mathbf v$ by taking $Df(0,0)\mathbf v$, which in this case would be $0$ for all vectors $\mathbf v$.
Thus, $f$ cannot be differentiable at $(0,0)$.
A: You may do it in spherical coordinates by setting $x=r \cos(\phi)$ and  $y=r \sin(\phi)$. Then the expression is equal to
$$\frac{1}{4}r \sin(4\phi)$$
-Mike
A: f will be differentiable at (0,0) $\iff$ 
$\displaystyle \lim_{(h,k)\rightarrow(0,0)}\frac{f(h,k)-f_{x}(0,0)h-f_{y}(0,0)k}{\sqrt{h^2+k^2}}=0$.
Since $f_{x}(0,0)=0$ and $f_y(0,0)=0$, we need to show that 
$\displaystyle \lim_{(h,k)\rightarrow(0,0)}\frac{f(h,k)}{\sqrt{h^2+k^2}}=\lim_{(h,k)\rightarrow(0,0)}\frac{hk(h^2-k^2)}{(h^2+k^2)^2}\ne0$; and 
this can be done by considering the limit along the line $h=2k$, say, or by using polar coordinates.
A: Your function can be plotted like this:

I plotted it by entering ((x)*y*(x*x - y*y))/((x*x + y*y)^(3/2)) + 4 to this website.
I would shift the function up or down $g(x,y) = f(x, y) + k)$. After that, proving $g(x,y)$ is not differentiable at $(k,k)$ is same as proving that $f(x,y)$ is not differentiable at $(0,0)$.
