# How does this prove the function has derivatives in all directions?

This is the function I'm analyzing: $$g(x,y)= \begin{cases} \dfrac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y)\neq(0,0)\\ 0 & \text{if } (x,y) = (0,0)\\ \end{cases}$$

I need to prove it has derivatives in all directions at $(0,0)$, so I applied the definition using a generic vector $v=(A,B)$:

$$\lim_{h\to 0} \frac {g(0+hA;0+hB)-g(0,0)}{h}$$

and I finally got to $A^3Bh-AB^3h$

Now I'm not sure what the conclusion is. Have I proven the function has derivatives in all directions at $(0,0)$?

Also, how can I tell a function doesn't have derivatives in all directions?

EDIT: corrected my result to add the missing $h$. However, my question remains the same: how does this prove the directional derivatives exist? What would a result be if they didn't exist?

Thanks.

You should be using $$\lim_{h \to 0}\frac{f(0+hA, 0+hB)-f(0,0)}{h\sqrt{A^2+B^2}}.$$
But the main thing is you forgot a factor of $h$ when evaluating $f(0+hA,0+hB)$. The ratio is \begin{align} \frac{h^2AB(A^2-B^2)/(A^2+B^2)}{h\sqrt{A^2+B^2}} &= h \frac{AB(A^2-B^2)}{(A^2+B^2)^{3/2}} \overset{h\to 0}{\longrightarrow} 0. \end{align}