How do I calculate the maximum value of the directional derivative of this function at the point $(0,0)$?
$$f(x,y)=\sqrt[3]{x^2y}$$
I did some calculations, but my answer came out to be $(0,0)$ and I can't figure out what I did wrong.
Basically, you do the partial derivative for the function,
\begin{align} f_x = \frac{2\sqrt[3]{y}} {3\sqrt[3]{x}} \qquad\qquad \text{and} \qquad\qquad f_y = \frac{\sqrt[3]{x^2}} {3\sqrt[3]{y^2}} \end{align}
Now if you use the point $(0,0)$ this becomes $0$, but it's not the right answer.
Forget about partial derivatives and work with the values $$D_\phi f(0,0):=\lim_{r\to0+}{f(r\cos\phi,r\sin\phi)-f(0,0)\over r}=\root 3\of {\cos^2\phi\sin\phi}\ .$$ When $\pi\leq\phi\leq2\pi$ the right hand side is $\leq0$ and drops out of consideration. In order to find the maximum of the right hand side in the interval $0\leq\phi\leq\pi$ we consider the function $$g(\phi):=\cos^2\phi\sin\phi=(1-\sin^2\phi)\sin\phi\qquad(0\leq\phi\leq\pi)\ .$$ Since the last expression depends only on $\sin\phi=:u$ we may as well look at $$h(u):=u-u^3\qquad(0\leq u\leq1)\ .$$ One has $h(0)=h(1)=0$ and $h'(u)=1-3u^2$. It follows that $h$ takes its maximum at $u_*=3^{-1/2}$, and the maximal value is $h(u_*)=2\cdot3^{-3/2}$. Let $\phi_*:=\arcsin u_*$. Then $g(\phi)$ is maximal when $\phi=\phi_*$ or $\phi=\pi-\phi_*$, and one has $$g(\phi_*)=h(u_*)=2\cdot3^{-3/2}\ .$$ The directional derivative $D_\phi f(0,0)$ is maximal for the same two values of $\phi$, and the maximal value is $$\max_\phi D_\phi f(0,0)=\bigl(g(\phi_*)\bigr)^{1/3}=\bigl(h(u_*)\bigr)^{1/3}={2^{1/3}\over 3^{1/2}}\ .$$