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: 
We consider    the function
  \begin{align*}
f(x,y)=\begin{cases}
\frac{x^4-y^3}{x^2+y^2}&\qquad\text{if }(x,y)\ne (0,0)\\
0&\qquad\text{if }(x,y)=(0,0)
\end{cases}
\end{align*}

As already stated by OP we see the partial derivatives exist.
\begin{align*}
\frac{\partial f}{\partial x}(0,0)&=\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h}
=\lim_{h\rightarrow 0}\frac{h^4}{h^3}=0\\
\frac{\partial f}{\partial y}(0,0)&=\lim_{h\rightarrow 0}\frac{f(0,h)-f(0,0)}{h}
=\lim_{h\rightarrow 0}\frac{-h^3}{h^3}=-1\\
\end{align*}

We observe that if the function $f$ is differentiable at $(0,0)$ the linear approximation $l(x,y)$ must be given as
  \begin{align*}
l(x,y)&=f(0,0)+\frac{\partial f}{\partial x}(0,0)\cdot x+\frac{\partial f}{\partial y}(0,0)\cdot y\\
&=-y\tag{1}
\end{align*}
For $f$ to be differentiable at $(0,0)$ we would need the limit
  \begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-l(x,y)}{\sqrt{x^2+y^2}}
\end{align*}
  to be equal to $0$ due to (1). Here the limit is
  \begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\frac{x^4-y^3}{(x^2+y^2)^{\frac{3}{2}}}
\end{align*}

If this limit exists, it must be equal to the limit as $(x,y)$ approaches the origin along the line $y=x$, which would be
\begin{align*}
\lim_{x\rightarrow 0}\frac{x^4-x^3}{(2x^2)^{\frac{3}{2}}}
=\frac{1}{2\sqrt{2}}\lim_{x\rightarrow 0}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}\tag{2}
\end{align*}

Since
  \begin{align*}
\lim_{x\rightarrow 0^+}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}=-1\qquad\qquad\text{resp.}\qquad\qquad
\lim_{x\rightarrow 0^-}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}=1\\
\end{align*}
  the limit (2) is not equal to $0$ showing the function $f$ is not differentiable at $(0,0)$.

A: 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.
A: Let $Q(x,y):=\frac{x^4+x^2y}{(x^2+y^2)^{\frac{3}{2}}}$. Then show that for $y=x>0$ we have
$Q(x,x)=\frac{1}{\sqrt{8}}(x+1) \to \frac{1}{\sqrt{8}} \ne 0$ for $x \to 0$.
