Prove this function is not differentiable For a function $f$ of $(x, y)$, $$f(x, y) = \begin{cases} 
          \frac{2x^2y}{x^4+y^2} & (x, y) \not= (0, 0) \\
          0 & (x, y) = (0, 0) \\
       \end{cases}
    $$
If the directional derivative of this function at $(0, 0)$ is $D_{\bar u}$, and given $D_\bar uf(0, 0)\not=\nabla f(0, 0) \bullet \bar u$, where $\bar u$ is a random unit vector $(u_1, u_2)$, use this to explain why $f$ is not differentiable at $(0, 0)$.
The solution I have is, first find $D_\bar u$. This value is $0$. Also $\nabla f$ is $(0, 0)$. So how can I use these values and $D_\bar uf(0, 0)\not=\nabla f(0, 0) \bullet \bar u$ to prove that $f$ is not differentiable? I know I can also show this is not a continuous function, but I want to prove it this way.

If $u = (u_1, u_2) \neq (0, 0)$, then
\begin{align*}
D_u f(0, 0) &= \lim_{t \to 0} \frac{f(tu) - f(0, 0)}{t} \\
&= \lim_{t \to 0} \frac{2(tu_1)^2(tu_2)}{t((tu_1)^4 + (tu_2)^2)} \\
&= \lim_{t \to 0} \frac{2t^3 u_1^2u_2}{t^5u_1^4 + t^3u_2^2} \\
&= \lim_{t \to 0} \frac{2u_1^2u_2}{t^2u_1^4 + u_2^2} \\
&= \frac{2u_1^2u_2}{u_2^2} = \frac{2u_1^2}{u_2},
\end{align*}
provided that $u_2 \neq 0$. This should linearly depend on $u$ if $f$ were differentiable at $0$, but it doesn't. There is no vector $v$ such that $D_uf(0, 0) = v \cdot u$. This shows that $f$ is not differentiable at $0$.
