<s>[INST] Directional derivatives at the origin and conditions for differentiability 
Let $f:\mathbb{R^2} \to \mathbb{R}$ $$f(x) = \left\{
\begin{array}{ll}
      \frac{xy^2}{x^2+y^4}, & (x,y)\ne0 \\
      0, & (x,y) =0 \\
\end{array} 
\right.$$ Find the directional derivatives at the origin $D_af(0,0)$ for every direction $a=(a_1,a_2)$, when $||a||=1.$ Show that $f$ is not differentiable at the origin.

For the partials I found that $\frac{\partial}{\partial x} = \frac{y^2(y^4-x^2)}{(x^2+y^4)^2}$ and $\frac{\partial}{\partial y} = \frac{2xy(x^2-y^4)}{(x^2+y^4)^2}$
so $\nabla f=(\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})$.
The directional derivative is then $D_af=\nabla f\cdot a = (\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})\cdot(a_1,a_2)$
I'm not sure I understand what they mean by $D_af(0,0) = \nabla f(0,0)\cdot a$ this would lead to division by $0$ right?
Also for the differentiability I tried to use the definition of the partials and see if they're both continuous at the origin, but that lead to a very messy expression for instance $$\frac{\partial}{\partial x} = \frac{f(x+h,y)-f(x,y)}{h} = \frac{\frac{(x+h)y^2}{(x+h)+y^4}-\frac{xy}{x^2+y^4}}{h}$$
and this didn't seem to simplify to anything usable... What should I do here?
 [/INST] We are requested to find the directional derivatives at the origin that is for $a\cdot b \neq 0$
$$\lim_{(ah,bh)\to(0,0)} \frac{\frac{ab^2h^3}{a^2h^2+b^4h^4}-0}{h} =\lim_{(ah,bh)\to(0,0)} \frac{ab^2h^3}{a^2h^3+b^4h^5}=\frac{b^2}a$$
with $f_x=f_y=0$.
For differentiability just note that for $x=y^2$
$$\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2+y^4}=\lim_{(y^2,y)\to(0,0)} \frac{y^4}{y^4+y^4}=\frac12$$
therefore $f(x,y)$ is not continuous at the origin and then it is not differentiable.
</s>