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Is it possible to a real-valued function of two variables defined on an open set to have partial derivatives of all order and to be discontinuous at some point or maybe at each point?

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In order that $\partial f/\partial x$ and $\partial f/\partial y$ be defined, $f$ would have to be continuous as a function of $x$ with $y$ fixed and vice-versa, so the discontinuity would have to happen as one approaches a point from a diagonal direction. –  Michael Hardy Apr 24 '12 at 17:29
If the partials exist and are continuous, then the function is differentiable, hence continuous at any point in the open set. –  copper.hat Apr 24 '12 at 19:30

3 Answers 3

$f(x,y)= 2xy/ x^2+y^2$ when $x^2+y^2>0$ and $0$ when $(x,y)=0$ f is differentiable at each point of other than the origin and $D_1f(0,0)=0=D_2f(0,0)$ since $f(x,0)=0=f(0,y) \forall x,y$ but But $f(t,t)=1 \forall t\neq 0$ so f is not continuous at $0$

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$f(x,y)=\frac{x^3y}{x^3+y^3}$ when $(x,y)\neq 0$ and $f(0,0)=0$

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Try $\frac {xy}{x^2 + y^2},(x,y) \ne (0,0), f(0,0) = 0$

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