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2 added 319 characters in body

I was travelling when this came up.

With the same results as Georges Elencwajg's comment and Marco's recent answer, take $$f(x,y) = \frac{2 x^2 y}{x^4 + y^2}$$ and set to $0$ at the origin $(0,0).$ Along any line through the origin $x = a t, \; y = b t$ the limit is 0, as $$| f(a t, b t) | \leq a^2 | t / b | .$$ However, along the parabola $y = x^2$ the value is 1, and along the parabola $y = - x^2$ the value is $-1.$

To get "directional derivative" 0 in every direction through the origin switch to $$g(x,y) = \frac{2 x^3 y}{x^6 + y^2}$$ as $$| g(a t, b t) | \; \leq \; t^2 \; | a^3 / b |$$ when $b \neq 0,$ but then if $y = \pm x^3$...the directional derivative generalizes to the Gateaux derivative in other settings.

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I was travelling when this came up. With the same results as Georges Elencwajg's comment and Marco's recent answer, take $$f(x,y) = \frac{2 x^2 y}{x^4 + y^2}$$ and set to $0$ at the origin $(0,0).$ Along any line through the origin $x = a t, \; y = b t$ the limit is 0, as $$| f(a t, b t) | \leq a^2 | t / b | .$$ However, along the parabola $y = x^2$ the value is 1, and along the parabola $y = - x^2$ the value is $-1.$