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I have to give an example of a function $f \colon \mathbb{R}^2 \to \mathbb{R}$ that has partial derivatives $(\partial_1f)(0,0) = 1 = (\partial_2f)(0,0)$ but no total derivative in $(0,0)$.

I think that the function $$ f \colon \mathbb{R}^2 \to \mathbb{R}, \quad f(x,y) = \begin{cases} \frac{xy}{x^2 + y^2} + x + y & \text{if $(x,y) \neq (0,0)$} \\ 0 & \text{if $(x,y) = (0,0)$} \end{cases} $$ is an example, I'm just not sure if my proof is correct.

We know that $f$ is total differentiable if $(x,y) \neq (0,0)$ because the partial derivatives exist and are continuous.

$f$ is also partial differentiable if $(x,y) = (0,0)$ because $f(x,0) = x$ and $f(0,y) = y$ and they are given by $(\partial_1 f)(x,0) = 1 = (\partial_2f)(0,y) = 1$.

To proof that $f$ is total differentiable in $(0,0)$ we now only have to proof that $$ f(x,y) = (1,1) \begin{pmatrix} x \\ y \\ \end{pmatrix} + o(\| (x,y) \|) \qquad \text{when $(x,y) \to (0,0)$} $$ but this doesn't hold when $x=y=t$ and $t \to 0$ and so $f$ is not total differentiable when $(x,y) = (0,0)$.

Can someone confirm if this is correct?

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    $\begingroup$ For me, it's correct. $\endgroup$ – paf Aug 26 '18 at 10:11
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    $\begingroup$ I found no error. $\endgroup$ – José Carlos Santos Aug 26 '18 at 10:13
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Yes, you have it right.

Your function has partial derivatives at the origin without being continuous.

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