Q:

The function $\frac{\sin(x^3+y^3)}{x^2+y^2}$ has all directional derivatives; is it differentiable at $(0,0)$?

Consider the function
$$f(x,y) = \frac {\sin(x^3+y^3)}{x^2+y^2}, \text{ for } (x,y)\neq (0,0),$$
with $f(0,0)=0$.
I have already proved that all directional derivatives exist.
Is the function differentiable at $(0,0)$?

A:

No. If it is differentiable then   the derivative is necessarily  given by $A(x,y)=x+y$. (By looking at the partial derivatives).  To get a contradiction from  this consider $\frac {f(x,y)-A(x,y)} {\sqrt {x^{2}+y^{2}}}=\frac {\sin (x^{3}+y^{3}) -(x+y)} {\sqrt {x^{2}+y^{2}}}$.  I will let you show that $\frac {\sin (x^{3}+y^{3})}  {\sqrt {x^{2}+y^{2}}} \to 0$ and $\frac {x+y} {\sqrt {x^{2}+y^{2}}} $does  not tend to $0$.

