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. 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 $f(x,y) = x^2y/(x^4+y^2)$. 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. 
I have verified the fact that the directional derivatives of $f$ and $g$ exist everywhere. Hints reqd to do the 2nd part.
 A: Let $u=(h,k)$, then $\lim_{t\to 0}\frac{h((0,0)+t(h,k))-h(0,0)}{t}=\lim_{t\to 0}\frac{h(th,tk)}{t}=\lim_{t\to 0}\frac{h^2(k+th^2)}{2t^2h^4+k^2+2tkh^2}$, with which, taking $u=(1,0)$ we realize that $\lim_{t\to 0}\frac{h^2(k+th^2)}{2t^2h^4+k^2+2tkh^2}=\lim_{t\to 0}\frac{1}{2t}$ and thus $h'(0,u)$ does not exist. This is also interpreted in another way by saying that the partial derivative with respect to the first variable in zero of $h$ does not exist and this is the same as saying that $h'(0,u)$ does not exist when $u=(1,0)$.
A: To start, use the chain rule. Let me know when you compute the matrix. I'm guessing it's a matter of looking at that and easily figuring out what $u$ should be. 
\begin{align*} h'(x_0,y_0) &= f'(g(x_0,y_0)) \cdot g'(x_0,y_0) \\ \\&= \begin{pmatrix} f_x  \\ f_y \end{pmatrix}_{g(x_0,y_0)} \cdot \begin{pmatrix} \frac{\partial (x)}{\partial x} & \frac{\partial (x)}{\partial y}  \\ \frac{\partial (y+x^2)}{\partial x} & \frac{\partial (y+x^2)}{\partial x}\end{pmatrix}_{(x_0,y_0)} \end{align*}
