Citing
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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.
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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?
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