# Show that the directional derivatives of $f$ and $g$ exist everywhere, but that there is a $u \neq 0$ for which $h'(0,u)$ does not

Let $g : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by the equation $g(x,y) = (x,y + x^2)$. Let $f : \mathbb{R}^2 \to \mathbb{R}$ be the function defined as $f(x,y) = x^2y/(x^4+y^2)$ if $(x,y)\neq (0,0)$ and $f(0,0)=(0,0)$. Let $h = f \circ g$.

Show that the directional derivatives of $f$ and $g$ exist everywhere, but that there is a $u \neq 0$ for which $h'(0,u)$ does not exist.

A part of the problem is already solved in this post Show that the directional derivatives of $f$ and $g$ exist everywhere, but that there is a $u \neq 0$ for which $h'(0,u)$ does not exist. but I still have doubts with the first part: How do I show that the directional derivatives of $f$ and $g$ exist everywhere? I have tried to use the definition and I come to the following:

If $(x,y)\neq (0,0), u=(h.k)$ then $\lim_{t\to 0}\frac{f(0+tu)-f(0)}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{h^2k}{t^2h^4+k^2}$, with which it would arrive at the condition that $hk=0$ to be able that the limit exists and thus to be able to say that $f'(0,u)$ exists, it is not supposed that there should not be conditions? Thank you very much

Your computation is off by one factor of $t$, and it ends up making a big difference.
$$\lim_{t\to0} \frac{x^2y}{x^4 + y^2} = \lim_{t\to0}\frac{(th)^2(tk)}{(th)^4+(tk)^2} = \lim_{t\to0}\frac{t^3h^2k}{t^4h^4+t^2k^2} = \lim_{t\to0}\frac{th^2k}{t^2h^4+k^2} = 0$$