Showing a function is not differentiable at $(0,0)$ 
Let $\displaystyle f(x,y)=\begin{cases} \frac{x^3+y^4}{x^2+y^2} \text{ if } (x,y) \neq (0,0)\\ 0 \text{ if } (x,y)=(0,0).  \end{cases}$  Show this is not differentiable at $(0,0)$.  

My strategy is to compare $D_uf(p)$ and $\nabla f(p)$ and hopefully $D_uf(p)\neq\nabla f(p)$.
So I computed the directional derivative:
Let $\displaystyle u=(u_1, u_2)$.  Then $\displaystyle D_uf(p)=\lim\limits_{t \to 0}\frac{f((0,0)+t(u_1,u_2))-f((0,0))}{t}=u_1^3$
"Work" 
$\displaystyle\frac{t^3u_1^3+t^4u_2^4}{t(t^2u_1^2+t^2u_2^2)}=\frac{u_1^3+tu_2}{u_1^2+u_2^2}=u_1^3+tu_2$
My problem is computing the gradient.  I worked it out by hand and got some stuff that looked like I would end up with $\frac{0}{0}$.  Is there another way to compute the gradient that I am missing.
 A: We want to show $f$ is not differentiable. We suppose to the contrary. Then $\nabla f$ exists, and $\nabla f\cdot u=D_u f$, for all $u$. That is, the tangent plane is well defined. If $\nabla f$ exists, than you have a steepest direction and a reverse, and all the other directions lead to various inclinations as if you were on a plane. But if $f$ is not differentiable, this is not the case. $D_uf$ may exist for all $u$, but it won't be predicted like a plane. 
If you graph it, like here: http://www.wolframalpha.com/input/?i=%28x%5E3%2By%5E4%29%2F%28x%5E2%2By%5E2%29+
then you see the plane is not well defined at the origin. There's too many slopes coming in. 
Let's continue with our proof by contradiction. If $\nabla f$ exists, then we have $
$$u_1^3=D_uf(0,0)=\nabla f(0,0)\cdot u=f_x u_1+f_y u_2.$$
But this is impossible for constants $f_x(0,0)$, $f_y(0,0)$. 
A: Let's use the direct definition of differentiability - that is, existence of a matrix, $\mathbf{J(x_a)}$, such that
$$
\lim_{\mathbf{h\to 0}} \frac{\mathbf{f(x_a+h)-f(x_a)-J(x_a)h}}{||\mathbf{h}||} = \mathbf{0}
$$
In our case, the matrix is actually a vector. Using radial coordinates, this is easier. We have $x=r\cos\theta$ and $y=r\sin\theta$, and so $x^2+y^2=r^2$. So
$$
f(r,\theta) = r\cos^3\theta+r^2\sin^4\theta
$$
Which can be confirmed to be accurate $\forall (r,\theta)$. Now, at $\mathbf{x_a=0}$, we have $r=0$, and $\mathbf{h=r}=(x,y)=(r\cos\theta,r\sin\theta)$. So we ask if there exists a matrix $\mathbf{J}$ such that
$$
\lim_{r\to 0} \frac{r\cos^3\theta+r^2\sin^4\theta-\mathbf{J(0)\cdot r}}{r} = 0
$$
Note that $\mathbf{J(0)}$ cannot depend on $r$ or $\theta$. If we let $\mathbf{J(0)}=(a,b)$, then
$$\begin{align}
\lim_{r\to 0} \frac{r\cos^3\theta+r^2\sin^4\theta-ar\cos\theta-br\sin\theta}{r} &=\lim_{r\to 0} \cos^3\theta+r\sin^4\theta-a\cos\theta-b\sin\theta\\
&=\cos^3\theta-a\cos\theta-b\sin\theta
\end{align}$$
From here, it is easy to see that no such values of $a$ and $b$ may exist such that the limit equals zero, and so the function is not differentiable.
A: The easiest way might be to show that it is not continuous at the origin (if indeed it is not). Just show that as you approach the origin from different directions, you end up with different functional values in the limit.
As well, you can calculate the Jacobian matrix. Its transpose is the gradient. You have
$$
\vec{\nabla}f(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{bmatrix}^T.
$$
