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 $u_y(0,0)$?
 A: The function $u$ clearly has partial derivatives at all points $\neq (0,0)$, since it is an elementary function in that open domain. At $(0,0)$ the partial derivatives would be 
$$u_x(0,0)=\lim_{x\to0}\frac{u(x,0)-u(0,0)}{x}=\lim_{x\to0}\frac{0}{x}=0$$
Likewise
$$u_y(0,0)=\lim_{y\to0}\frac{u(0,y)-u(0,0)}{y}=\lim_{y\to0}\frac{0}{y}=0$$
Therefore, the partial derivatives exist everywhere.
It looks then, that the point of the exercise is to show that differentiability and existence of partial derivatives are not equivalent. This rounds up the relation between these two concepts, since differentiability at a point, implies that the partial derivatives exist.
A: Having partial derivatives at every point is not the same as being differentiable.
The usual formulas, i.e. in this case the quotient rule and the product rule, show that $(x,y) \mapsto \dfrac{xy}{x^2+y^2}$ has partial derivatives with respect to $x$ and $y$ at every point except the one point where the numerator and denominator are both $0.$ At that point, the existence of partial derivatives can be shown by applying the definition of differentiation:
$$
\left.\frac{\partial u}{\partial x}\right|_{(x,y)\,=\,(0,0)} =\lim_{\Delta x\,\to\,0} \frac{u(0+\Delta x, 0) - u(0,0)}{\Delta x} = \lim_{\Delta x\,\to\,0} \frac{{0} - 0}{\Delta x} = 0
$$
and similarly for the other one.
But $u$ is not differentiable at $(0,0)$ because there is no tangent plane. Since the partial derivatives with respect to $x$ and $y$ at $(0,0)$ are both $0,$ if there is a tangent plane its equation must be $z = 0 + 0(x-0) + 0(y-0).$
However, that will not work for the following reason. Suppose one moves away from $(0,0)$ in a direction at a $45^\circ$ angle to the two coordinate axes. That is the line $x=y.$ If $x=y$, then $u(x,y) = u(x,x) = \dfrac{2x^2}{x^2+x^2} = 1.$ This is not even continuous at $(0,0)$, and if it were, if the derivative in that direction were anything but $0,$ then it would not fit the equation of the tangent plane just given.
A: "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!
