Could someone please assist with the following questions:
Consider $f(x,y) = x^{\frac{1}{3}}y^{\frac{1}{3}}$ and take $C$ to be the curve of intersection of $z = f(x,y)$ with the plane $y=x$. Show that the curve $C$ has a tangent line at the origin?
I have tried showing that the directional derivative $D_{u}f(0,0)$ exists (using the formula $D_{u}f(x_{0},y_{0}) = \lim\limits_{h \rightarrow 0}\frac{f(x_{0}+ha,y_{0}+hb)-f(x_{0},y_{0})}{h}$) by taking $u = \frac{(1,1)}{\sqrt{2}}$ (since $y = x$), but I get a limit which does not exist.
Lastly, why is $f$ not differentiable at $(0,0)$?
Any assistance would be appreciated.
The curve has a cusp at the origin, but each of the two branches emanating there has a "onesided" tangent, namely the positive $z$-axis.
Introduce an $s$-axis with positive direction $\bigl({1\over\sqrt{2}},{1\over\sqrt{2}},0\bigr)$. Your curve is lying in the $(s,z)$-plane, and in this plane it is described by the equation $$z^3={s^2\over2}\ .$$ The two branches can then be written as $$s=s_\pm(z):=\pm \sqrt{2} \>z^{3/2}\qquad(z\geq0)\ ,$$ and it is easy to see that $s'_\pm(0)=0$.
The curve $C$ is a one-dimensional object in three-space. Its two branches $C_\pm$ are therefore best presented in terms of a parametric representation $$C_\pm:\quad t\mapsto {\bf r}(t):=\bigl(\pm t^{3/2},\pm t^{3/2},t\bigr)\qquad(0\leq t<\infty)\ .$$ The tangent vector at $(0,0,0)$ is then given by ${\bf r}'(0)=(0,0,1)$, and is the same for both branches. By the way: The curvature of $C_\pm$ is $\infty$ at the origin.
It is true that at each point of $C$ one has $z^3=x^2$, or $z^3=y^2$. But it would be misleading to consider $z^3=x^2$ as equation of the curve. Interpreted in $(x,y,z)$-space the equation $z^3=x^2$ defines an infinite surface in the shape of a "paper double wing".