Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x)=\begin{cases} \frac{x_1^2x_2}{x_1^2+x_2^2}, \quad x\neq 0 \\ 0 , \quad \quad \, \, \, \, x=0\end {cases}$.
I suppose that the derivative doesn't exist, but how can I prove that $f$ is not differentiable at $0$? I tried using $\lim\limits_{h\to 0}\frac{f(a+h)-f(a)-Lh}{|h|_2}$ but how can I show that there is no linear function $L$?
Both partial derivatives are $0$ at the origin. If $f$ were differentiable there, then we would have
$$\tag 1 f(x,y) = f(0,0) + f_x(0,0)\cdot x + f_y(0,0)\cdot y + o(\sqrt {x^2+y^2}\,)$$
as $(x,y)\to (0,0).$ But note the right side of $(1)$ is just $o(\sqrt {x^2+y^2}\,).$ Now observe $f(x,x)= x,$ which is not $o(\sqrt {x^2+x^2})$ as $x\to 0,$
Aiming for a contradiction, assume f is differentiable and choose $ h_1 = t $ and $h_2=t$ and show that $\lim_{(t,t) \to (0,0)}\frac{f(t,t)}{\Vert(t,t)\Vert} = \frac{1}{2\sqrt{2}}\neq 0$.
You know that $Lh = 0$ and $f(0,0)=0$. Therefore $f(h_1,h_2)$ equals the error in the approximation which should be $0$ for a sufficient small $h$. But with our choice of $h$ this doesn't happen which leads to a contradiction that $f$ is differentiable in $(0,0)$.
