Consider $f(x,y) = \begin{cases}\frac {x^3+y^3}{x^2+y^2}&(x,y)\neq (0,0)\\0& (x,y)=(0,0)\end{cases}$. Does the total derivative at the origin exist?
I checked that the function is continuous at the origin but the partial derivatives are not continuous. So, I am not sure how to proceed.
Since $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=1$, if $f$ is differentiable at $(0,0)$, its derivative at $(0,0)$ is the linear map $(x,y)\mapsto x+y$. Now, since$$\frac{x^3+y^3}{x^2+y^2}-f(0,0)-(x+y)=\frac{-x^2y-xy^2}{x^2+y^2}=-\frac{xy(x+y)}{x^2+y^2},$$the question is: is it true that$$\lim_{(x,y)\to(0,0)}\frac{\left|-\frac{xy(x+y)}{x^2+y^2}\right|}{\|(x,y)\|}=0?$$No. See what happens whe $x=y$.