# Differentiability of a function where all directional derivatives exist

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

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 $$0$$.