Differentiability in $(0,0)$ of $f(x,y)=\frac{x^4-y^3}{x^2+y^2}$ Let the function $f$ be defined as $$f(x,y)=\frac{x^4-y^3}{x^2+y^2}$$
If $(x,y)=(0,0)$ then $f$ is equal to zero.
My problem is to prove that this function isn't differentiable at the point $(0,0)$.
My solution:
 First idea is maybe to show that $f$ is not continuous at $(0,0)$. That will show that $f$ isn't differentiable at $(0,0)$.
 But it is easy to see that $f$ is continuous at $(0,0)$, so we need to check the partial derivatives at $(0,0)$.
 Partial derivative in $x$-direction is
$$\lim_{h\to 0}\frac{f(h,0)-0}{h}=0$$
 Partial derivative in $y$-direction is $$\lim_{h\to 0}\frac{f(0,h)-0}{h}=-1$$
We now know how our Jacobian matrix at the point $(0,0)$ looks. We now use the definition of derivatives:
$$\lim_{(x,y)\rightarrow(0,0)}\frac{\frac{x^4-y^3}{x^2+y^2}+y}{\sqrt{(x^2+y^2)}}=\frac{x^4+x^2y}{(x^2+y^2)^{\frac{3}{2}}}$$
But this looks like having a limit in $(0,0)$ and it's equal to $0$? 

A necessary condition for being differentiable at $(0,0)$ is that the directional derivative: 
$$ Df_{(0,0)}(h,k) = \lim_{t\rightarrow 0} \frac{1}{t} (f(th,tk)-f(0,0)) = 
 \lim_{t\rightarrow 0} \frac{t^4 h^4 - t^3 k^3}{t^3(h^2+k^2)} = \frac{-k^3}{h^2+k^2}$$
should be a linear function of $h$ and $k$ (i.e. of the form $ah+bk$ for real constants $a$ and $b$), which is visibly not the case. In your own solution you may simply calculate $Df_{(0,0)} (1,1)$ and show that it is not the sum of $Df_{(0,0)} (1,0)$ and $Df_{(0,0)} (0,1)$, so it is not differentiable.
