Direction where the directional derivative is maximal I am given the following function:
$$
f(x,y)=\sqrt[3]{x^2 y }
$$
at (0,0),
and need to find the directions $\vec{v}$ for which the directional derivative $D_\vec{v} f (0,0)$  is maximal.
I know the answer should be $ (\pm \sqrt{\frac{2}{3}}, \frac{1}{\sqrt{3}}) $.
Since this function is not even differentiable (I think), I should differentiate according to the definition:
$$
D_\vec{v} f (0,0) = \lim_{h\to 0} \frac{\sqrt[3]{\cos^2 (\theta) \sin(\theta )} h}{h} = \sqrt[3]{\cos^2 (\theta) \sin(\theta )} $$ and it seems like the only thing I should do now is to find the maximum value of $\sqrt[3]{x^2 y }$. Is there any simple way to do it?
Will you please help me?
 A: The directional derivative at a point $(x,y)$ and along the direction $\hat r=\hat x\cos \theta+\hat y \sin \theta$, where $\theta$ is the angle the unit vector $\hat r$ makes with the $x$-axis, is given by 
$$\lim_{h\to 0}\frac{f(x+\cos \theta h,y+\sin \theta h)-f(x,y)}{h}$$
Let $f$ be the function defined by $f(x,y)=x^{2/3}y^{1/2}$.  Then, the directional derivative at $(0,0)$ along $\hat r=\hat x\cos \theta+\hat y \sin \theta$ is 
$$\lim_{h\to 0}\frac{(h\cos \theta)^{2/3}(h\sin \theta)^{1/3}}{h}=\cos^{2/3}\theta \sin^{1/3}\theta$$
Local extrema of the directional derivative occur at values of $\theta$ where its derivative is zero.  Taking a derivative as setting to zero reveals
$$\begin{align}
\frac{d(\cos^{2/3}\theta \sin^{1/3}\theta)}{d\theta}&=-\frac23 \cos^{-1/3}\theta \sin^{4/3}\theta +\frac13\sin^{-2/3}\theta \cos^{5/3}\theta\\\\
&=0\implies \tan \theta =\pm \frac{\sqrt{2}}{2}
\end{align}$$
Since we seek the maximum directional derivative, we exclude those directions for which $\sin \theta<0$.  Therefore, the directional derivative is maximum along
$$\theta =\arctan\left(\frac{\sqrt{2}}{2}\right)$$
$$\theta =\pi -\arctan\left(\frac{\sqrt{2}}{2}\right)$$
which gives for $\hat r$
$$\bbox[5px,border:2px solid #C0A000]{\hat r=\pm \hat x\frac{\sqrt{6}}{3}+\hat y \frac{\sqrt{3}}{3}}$$
