Prove $f: \mathbb{R}^{2} \to \mathbb{R}$ has directional derivatives in $(0,0)$ but is not differentiable at $(0,0)$. Let
$$f(x,y) = \begin{cases} \frac{2x^2y}{x^{2}+y^{2}}, & \mbox{if } (x,y) \neq (0,0) ,  \\ 0, & \mbox{if } (x,y) =(0,0)\end{cases}$$
be a function from $\mathbb{R}^{2}$ to $\mathbb{R}$. Prove that in $(0,0)$ the function $f$ has directional derivatives in all directions. But $f$ is not differentiable in $(0,0)$. For the first question  I try to show that for $v=(x,y)$ and $t \in \mathbb{R}$:
$$D_{v}(f)=lim_{t \to 0} \frac{f(0+tv)-f(0)}{t}$$.
exist. This proves directional derivatives that in $(0,0)$ directional derivatives exist in any direction right? But I cant prove this limit is $0$. So far:
$$\frac{f(0+tv)-f(0)}{t}=\frac{f(tv)}{t}=\frac{f(tx,ty)}{t}=\frac{\frac{2(tx)^{2}(ty)}{(tx)^{2}+(ty)^{2}}}{t}=\frac{2x^{2}y}{x^{2}+y^{2}}.$$
Then $D_{v}(f)=lim_{t \to 0} \frac{f(0+tv)-f(0)}{t}$ exist and is equal to the constant $\frac{2x^{2}y}{x^{2}+y^{2}}$. This proves directional derivatives exist in $(0,0)$ in all directions. In order to show that $f$ is not differentiable in $(0,0)$, I attempted to prove $f$ is not continuous at $(0,0)$. But cant find the path to prove this, for instance $$lim_{t \to 0}f(t,t)=lim_{t \to 0} t=0=f(0,0)$$ also tried $$lim_{t \to 0}f(t^{3},t)=lim_{t \to 0} t^{3}=0=f(0,0)$$.
 A: Let $f(x,y)$ be defined as
$$f(x,y)=\begin{cases}\frac{x^2y}{x^2+y^2}, &(x,y)\ne(0,0)\\\\0,& (x,y)=(0,0)\end{cases}$$
The directional derivative of $f(x,y)$ along the unit vector $(a,b)$ (i.e. $a^2+b^2=1$) at $(x,y)=(0,0)$ is given by the limit
$$\begin{align}
\lim_{h\to 0}\frac{f(ah,bh)-f(0,0)}{h}&=2\lim_{h\to 0} \frac{\frac{a^2h^2bh}{a^2h^2+b^2h^2}}{h}\\\\
&=\frac{2a^2b}{a^2+b^2}\\\\
&=2a^2b
\end{align}$$
So, the directional derivatives exist in all directions at the origin.

It is easy to show that both first partial derivatives vanish at the origin.  It remains to show that the limit
$$\lim_{(h,k)\to (0,0)}\frac{f(h,k)-f(0,0)-f_1(0,0)h-f_2(0,0)k}{\sqrt{h^2+k^2}}$$
fails to exist.  Proceeding we see that
$$\begin{align}
\frac{f(h,k)-f(0,0)-f_1(0,0)h-f_2(0,0)k}{\sqrt{h^2+k^2}}&=\frac{2h^2k}{(h^2+k^2)^{3/2}}\\\\
&=2\frac{h^2}{h^2+k^2}\frac{k}{\sqrt{h^2+k^2}}\tag1
\end{align}$$
Now, if $h=k$, the limit of $(1)$ is $2^{-1/2}$ while if $h=0$ or $k=0$ the limit is $0$.  Inasmuch as the limit of $(1)$ fails to exist, we conclude that $f$ is not differentiable.
