Directional derivatives exercise from Courant's introduction to calculus and analysis Show for $z=f(x,y)=\sqrt[3]{xy}$ that $f$ is continuous and that the partial derivatives $\partial z/\partial x$ and $\partial z/\partial y$ exist at the origin but that the directional derivatives in all other directions do not exist.
I know that $\partial z/\partial x=\frac{y}{3(xy)^{2/3}}$. How to prove that $\partial z/\partial x$ exists at the origin (i.e. that $\lim\limits_{(x,y)\rightarrow(0,0)}\partial z/\partial x$ exists).
Moreover, if $(\partial z/\partial x)_{(0,0)}$ and $(\partial z/\partial y)_{(0,0)}$ exists, how could the directional derivatives $$D_{(\alpha)}f(x,y)=\frac{\partial f}{\partial x}\cos\alpha+\frac{\partial f}{\partial y}\sin\alpha$$ not exist?
That's exercise 5 of 1.5b
 A: Your formula for $D_{(\alpha)}f(0,0)$ is valid only if $f$ is indeed differentiable at $(0,0)$. But this is not the case here, even though the partial derivatives $f_x(0,0)$ and $f_y(0,0)$ exist.
Since $(0,0)$ is a very special point for $f$ we have to resort to the (geometric) definition of $D_{(\alpha)}f(0,0)$, namely
$$\eqalign{D_{(\alpha)}f(0,0)&:=\lim_{r\to0+}{f(0+r\cos\alpha,0+r\sin\alpha)-f(0,0)\over r}\cr &=\lim_{r\to0+}r^{-1/3}{\root 3 \of{\mathstrut\cos\alpha\sin\alpha}}\ .\cr}$$
Now the latter limit obviously does not exist when $\alpha$ is not an integer multple of ${\pi\over2}$.
A: 
Real valued  $z=f(x,y)=\sqrt[3]{xy}$ itself does not exist in the 2nd and 4th quadrants.
The origin is a singular point. Partial derivatives $\partial z/\partial x$ and $\partial z/\partial y$ exist only in these quadrant directions, including at the origin if approached along these quadrant directions. But $(\partial z/\partial x)_{(0,0)}$ does not  exist unconditionally.
Situation is pretty much the same for $z=f(x,y)=\sqrt{xy}$.
