Partial derivatives, directional derivatives and differentiability of a function I'm studying the function $f:\mathbb{R}^2\to \mathbb{R}$ such that:
$$f(x,y) = \begin{cases}
                  y & \text{if } x=0
                  \\
                  x^2+x^2\sin(\frac{1}{x}) & \text{if } x\neq 0
                  \end{cases}$$
How to find the partial derivatives and directional derivatives at $(0,0)$? Do they exist? 
Thanks for your help. 
 A: You find partial derivatives the usual way: the parial derivative with respect to $y$ at $(0,0)$ is
$$\frac{\partial f}{\partial y}\Biggm|_{(x,y)=(0,0)} = \lim_{y\to 0}\frac{f(0,y) - f(0,0)}{y-0} = \lim_{y\to 0}\frac{y-0}{y-0} = 1.$$
While the partial derivative with respect to $x$ at $(0,0)$ is:
$$\frac{\partial f}{\partial x}\Biggm|_{(x,y)=(0,0)} = \lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{x^2+x^2\sin(1/x) - 0}{x-0} = \lim_{x\to 0}\left(x+x\sin\frac{1}{x}\right)=0.$$
Added. I had missed the "directional derivative" part.
The directional derivatives are similar: let $\mathbf{u}=(a,b)$ be a unit vector. Then by definition we have that
$$D_{\mathbf{u}}f(0,0) = \lim_{h\to 0}\frac{f(ha,hb) - f(0,0)}{h}.$$
If $a=0$, you are looking at the partial with respect to $y$, which we have already computed. If $a\neq 0$, then you get
$$D_{\mathbf{u}}f(0,0) = \lim_{h\to 0}\frac{1}{h}\left(h^2a^2 + h^2a^2\sin(1/h^2a^2)\right) = \lim_{h\to 0}\left(ha^2 + ha^2\sin(1/h^2a^2)\right)=0.$$
