Let $f(x,y)=\frac{x^3-y^3}{x^2+y^2}$. Is f differentiable in $(0,0)$?

Question

Let $$f(x,y)=\frac{x^3-y^3}{x^2+y^2}$$

My solution manual says that this function is not diffb. in $(0,0)$ because it is not linear. Well my problem is that I don't see why this function is linear, and I also don't see why that would imply that $f$ is not differentiable.

Not sure about differentiability, but it's clear that $f$ is a linear function from $\mathbb{R}^2$ to $\mathbb{R}$ if we define $f(0, 0) = 0$. Just compare $f(kx, ky)$ to $f(x, y)$. — Hew Wolff, Apr 10 '13 at 19:22
@HewWolff: How the homogeneity of $f(x,y)$ could help the OP? — Mikasa, Apr 10 '13 at 19:39
@HewWolff The function is not linear, just homogeneous of degree $1$. — Pedro, May 15 '13 at 15:54

score 5 · Answer 1 · answered May 15 '13 at 15:53

We find the directional derivatives at $(0,0)$. Assume $\Vert v\Vert =1$. Then

$$f'((0,0);v)=\lim_{h\to 0}\frac{f((0,0)+h\cdot v)-f(0,0)}{h}$$

I take $f(0,0)=0$ to ensure continuity.

$$f'((0,0);v)=\lim_{h\to 0}\frac{f(hv_1,hv_2)}{h}$$

$$f'((0,0);v)=\lim_{h\to 0}\frac{h^3v_1^3-h^3v_2^3}{h^3}\frac{1}{\Vert v\Vert}$$

$$f'((0,0);v)=\lim_{h\to 0}\frac{h^3v_1^3-h^3v_2^3}{h^3} $$

$$f'((0,0);v)=v_1^3-v_2^3$$

Since the directional derivatives do not follow a linear dependence on $(v_1,v_2)$, we conclude $f$ cannot be differentiable at $(0,0)$.

Let $f(x,y)=\frac{x^3-y^3}{x^2+y^2}$. Is f differentiable in $(0,0)$?

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