Sanity check on example 6.5 from "Counterexamples in probability and real analysis" by Wise and Hall Citing
"
Example 6.5. There exists a real-valued function defined on the plane that is discontinuous and yet such that both partial derivatives exist and are continuous.
Proof It is well know that if $f:\mathbb{R} \to \mathbb{R}$ is differentiable, then $f$ is continuous. What if $g:\mathbb{R}^2 \to \mathbb{R}$ is such that both partial derivatives exist and are continuous? Must $g$ be continuous? Consider the function defined via
$$g(x,y)= \begin{cases} \frac{xy}{x^2+y^2} & \text{ if } (x,y) \neq (0,0) \\ 0 &\text{ if } (x,y)=(0,0)\end{cases}$$
Then it trivially follows that each partial derivative of g exists and is continuous. However, note that $g$ is not continuous. Note also that this function g is continuous in each variable but is not continuous.
"
However, computing the $x$ partial derivative at $(x,y) \neq (0,0)$ we get, for $x=0,y \neq 0, 1/y$. So the partial derivatives cannot be continuous at $(x,y)=(0,0)$ even though they exist there.
Is it, as I think, some elementary mistake by the authors, or is it me who don't understand something obvious?
If it is indeed such an elementary mistake, is this book considered in general to be otherwise trustable?
 A: So, throwing away political correctness: yes, there is an elementary mistake in the text. I suppose that it should have been something like:
"
Example 6.5. There exists a real-valued function defined on the plane that is discontinuous and yet such that both partial derivatives exist everywhere in the plane.
Proof It is well known that if $f:\mathbb{R} \to \mathbb{R}$ is differentiable everywhere, then $f$ is continuous. What if $g:\mathbb{R}^2 \to \mathbb{R}$ is such that both partial derivatives exist everywhere? Must $g$ be continuous? Consider the function defined via
$$g(x,y)= \begin{cases} \frac{xy}{x^2+y^2} & \text{ if } (x,y) \neq (0,0) \\ 0 &\text{ if } (x,y)=(0,0)\end{cases}$$
Then each partial derivative of g exists at every point of the plane. However, note that $g$ is not continuous. Note also that this function g is continuous in each variable but is not continuous.
"
The book "Counterexamples in Analysis" by Gelbaum and Olmsted does not make this mistake (9.4).
A: Your direct argument that the partial derivatives are not continuous at $(0,0)$ is  correct.  
To confirm, although the partial derivatives at $(0,0)$ exist, with
$$\frac{\partial g}{\partial x}(0,0) = \frac{\partial g}{\partial y}(0,0) =0 $$
they cannot be continuous in an open neighborhood of $(0,0).$ Otherwise the derivative at $(0,0)$
$$Dg(0,0) : (h,k) \mapsto Dg(0,0)(h,k),$$
must exist as the unique linear operator where
$$Dg(0,0)(h,k) = h\frac{\partial g}{\partial x}(0,0) + k\frac{\partial g}{\partial y}(0,0) = 0,$$
and where we have existence of a finite value for 
$$\lim_{(h,k) \to (0,0)} \frac{g(h,k) - g(0,0) - Dg(0,0)(h,k)}{\sqrt{h^2 + k^2}}= \lim_{(h,k) \to (0,0)} \frac{hk}{(h^2 + k^2) \sqrt{h^2 + k^2}}.$$
This is impossible since taking $h = k > 0$ we have
$$\lim_{(h,k) \to (0,0)} \frac{hk}{(h^2 + k^2) \sqrt{h^2 + k^2}} = \lim_{h \to 0+} \frac{h^2}{2\sqrt{2}h^3}  = +\infty$$
A: As is well known, if both $\partial f/\partial x, \partial f/\partial y$ exist everywhere and are continuous functions on the plane, then $f$ is differentiable everywhere, hence $f$ is continuous everywhere.
