Given that $$f(x,y)= \begin{cases} \frac{xy}{x^2+y^2}, & \text{if $(x,y)\neq0$}.\\ 0, & \text{if $(x,y)=0$} \end{cases}$$ I want to show that $f$ cannot be totally differentiable at $(0,0)$.
I can show that $f$ is not totally differentiable at $(0,0)$ by showing that it isnt continous at $(0,0)$, however I need to prove it using the definition of total differentiability, which says that some $f$ is totally differentiable at $x$ if there exists some $L$ such that $$f(x+\xi)=f(x)+L\xi+\varphi(\xi)$$ where $$\lim_{\xi \to 0} \frac{\varphi(\xi)}{\ \|\xi\|}=0$$ I am confused by the notation of this definition. Surely this definition should be based around the idea of differentiability at $(x,y)$, not just $x$?
$$f(cy,y)=\frac{cy^2}{cy^2+y^2}=\frac{c}{c+1}$$ So $$\lim_{\xi\to 0}f(\xi)\neq 0$$
Now assume such an $L$ exists. Then:
$$||f(x+\xi)-f(x)||=||L\xi+\varphi(\xi)||\leq||L||\;||\xi||+||\varphi(\xi)||$$
Since $\lim_{\xi\to 0} \frac{||\varphi(\xi)||}{||\xi||}=0$ we find a $\delta>0$ such that for all $\xi\in B(0;\delta)$ we have $||\varphi(\xi)||/||\xi||\leq 1$. Now: $$||\varphi(\xi)||\leq ||\xi||$$ Combining with the previous result we see: $$||f(x+\xi)-f(x)||\leq(||L||+1)||\xi||$$ Hence $||f(x+\xi)-f(x)||\to 0 $ as $\xi\to 0$.
So we see that: $$\lim_{\xi\to 0} f(\xi)= 0$$
Contradicting the previous claim.
Basically what we have done here is prove that when $f$ is totally differentiable in a point, then it is continuous in that point, and use the fact that $f$ is not continuous in $0$. Sadly I don't see another way to do this.