f(x,y) not Differentiable on (0,0)

Suppose that $f: \mathbb{R}^2\to \mathbb{R}$ is constructed to be continuous s.t.

i) $f(x,y)= 0$ unless $x>0$ and $x^2<y<3x^2$

ii) for each $x>0$, $f(x,2x^2)=x$

iii) $0 \leq f(x,y) \leq x$, for all $(x,y)$ with $x>0$.

Show all directional derivatives of $f$ vanish at $(0,0)$ Show $f$ is not differentiable at $(0,0)$.

The first part is easy and I proved it. The second one troubles me. I considered taking the limit of the parabola $\{(t,t^2): t>0\}$ and showing that it doesn't exist. Am I on the right track? Help? (it's from an old past exam paper w/ no solutions :/)

Solution 1:

We have $f_x(0,0)=0=f_y(0,0)$ and $f(0,0)=0.$

Suppose that $f$ is differentiable at $(0,0)$. Then

$$g(x,y):=\frac{f(x,y)}{\sqrt{x^2+y^2}} \to 0$$

as $(x,y) \to (0,0)$.

But this is not the case, since for $x>0$ and $y=2x^2$ we have

$$g(x,2x^2)= \frac{1}{\sqrt{1+4x^2}} \to 1$$

as $x \to 0.$