What is the meaning of "has partial derivatives everywhere"? I am reading Courant's "Differential and Integral Calculus Vol.2".
I am confused with something: Courant says that the function defined as $u(x,y)=\frac{2xy}{x^2 +y^2}$ and $u (0,0)=0$ is not continuous but has partial derivatives everywhere.

I am assuming that "has partial derivatives everywhere" means "is differentiable". I could be wrong.

I've seen somewhere else that a function is differentiable at $(x_0, y_0)$ if there is $(\alpha_1, \alpha_2)\in \mathbb{R}^2$ such that:
$$\lim_{(h,k)\to (0,0)} \frac{f(x_0+ h, y_0  + k )-f(x_0,y_0) - \alpha_1 h - \alpha_2 k }{\sqrt{h^2 + k^2}}=0$$  
Applying to our function, we have:
$$\lim_{(h,k)\to (0,0)} \frac{2hk}{(h^2 + k^2)\sqrt{h^2 + k^2}}-\frac{\alpha_1 h +\alpha_2 k}{\sqrt{h^2 + k^2}}$$
With polar coordinates $h\to r\cos, k\to r \sin$. 
$$\lim_{r\to 0} \frac{2\cos \theta \sin \theta}{r}- \alpha_1 \cos \theta - \alpha_2 \sin \theta = \lim_{r\to 0} \frac{\sin{2\theta}}{r}  - \alpha_1 \cos \theta - \alpha_2 \sin \theta $$
I guess this shows that the limit depends on $\sin 2\theta$ being equal to $0$ and hence can not exist. So what Courant meant with "partial derivatives existing everywhere"? It could be $\mathbb{R}\setminus\{(0,0)\}$ but I wouldn't call this "everywhere".
Also, when defining $u(0,0)=0$, what does this means for the partial derivatives? What does this means for $u_x(0,0)$ and $u_y(0,0)$?

"Has partial derivatives everywhere" means quite literally that it has partial derivatives everywhere.  In other words, at every point, the partial derivatives of $u$ with respect to each variable exist.  In other words, for all $(x,y)\in\mathbb{R}^2$ the limits $$u_x(x,y)=\lim_{h\to 0}\frac{u(x+h,y)-u(x,y)}{h}$$ and $$u_y(x,y)=\lim_{h\to 0}\frac{u(x,y+h)-u(x,y)}{h}$$ exist.
So, this doesn't have anything at all to do with $u$ being differentiable in your sense, at least not directly.  You are correct that $u$ is not differentiable at $(0,0)$ (it is not even continuous there!), but the partial derivatives $u_x(0,0)$ and $u_y(0,0)$ still exist.  Try computing them directly from the limit definitions above!
