Second partial derivatives vs total second derivative Suppose that second partial derivatives exist at $(a,b)$ and $\dfrac{\partial^2f}{\partial x\,\partial y}(a,b)\neq\dfrac{\partial^2f}{\partial y\,\partial x}(a,b)$.
(For example $f(x,y)=\begin{cases}xy\frac{x^2-y^2}{x^2+y^2}\quad\textrm{for } (x,y)\neq (0,0)\\0,\hspace{16mm}\textrm{for } (x,y)=(0,0)\end{cases} $)
Can function $f$ be twice differentiable at $(a,b)$ (The second-order total differential exist)? 
Thank you.  

Great question. So consider the function $F\colon\mathbb R^2\to L(\mathbb R^2,\mathbb R)\cong\mathbb R^2$ given by $F(x,y) = df(x,y)$. We have $F(0,0)=0$. With a little work (or Mathematica) we find that 
$$F(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x}(x,y) \\ \frac{\partial f}{\partial y}(x,y) \end{bmatrix} = \begin{bmatrix} \frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \\\frac{x(x^4-4x^2y^2-y^4)}{(x^2+y^2)^2}\end{bmatrix}\,.$$
If $F$ is differentiable at $(0,0)$, of course its derivative must be given—as a bilinear form—by the $2\times 2$ matrix
$$dF(0,0)=\begin{bmatrix} \tfrac{\partial^2 f}{\partial x^2}(0,0) & \tfrac{\partial^2 f}{\partial y\partial x}(0,0) \\ \tfrac{\partial^2 f}{\partial x\partial y}(0,0) & \tfrac{\partial^2 f}{\partial y^2}(0,0) \end{bmatrix} = \begin{bmatrix} 0&-1\\1&0\end{bmatrix}\,.$$
Now, if $F$ is differentiable at $(0,0)$, we must have
$$\lim_{(x,y)\to (0,0)} \frac{F(x,y)-F(0,0)-dF(0,0)(x,y)}{\sqrt{x^2+y^2}} = 0\,.$$
Try it along the line $y=x$. The numerator becomes
$$\begin{bmatrix} x\\-x \end{bmatrix} - \begin{bmatrix} -x\\x \end{bmatrix} = 2\begin{bmatrix} x\\-x \end{bmatrix}\,,$$
and 
$$\lim_{x\to 0} \frac{2 \begin{bmatrix} x\\-x \end{bmatrix}}{\sqrt 2|x|}$$
certainly is not $0$. Thus, $F$ is not differentiable at $(0,0)$.
