Vtyjkyuk

I have to give an example of a function $f colon mathbbR^2 to mathbbR$ 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
$$
f colon mathbbR^2 to mathbbR,
quad
f(x,y) =
begincases
fracxyx^2 + y^2 + x + y & textif $(x,y) neq (0,0)$ \
0 & textif $(x,y) = (0,0)$
endcases
$$
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
$$
f(x,y)
= (1,1)
beginpmatrix
x \
y \
endpmatrix
+ o(| (x,y) |)
qquad
textwhen $(x,y) to (0,0)$
$$
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?

Yes, you have it right.

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