# Partial derivatives of $f(x,y)=\frac{\sin(x^3+y^3)}{x^2+y^2}$ in (0,0) where f(0,0)=0

Let f be defined as: $$f(x,y)=\frac{\sin(x^3+y^3)}{x^2+y^2}$$ and $$f(0,0)=0$$.

a) For which directions $$v \in \mathbb{R}^2\setminus\{0\}$$ is there a directional derivative $$d_vf(0,0)$$. Calculate it in these cases.

b) Is $$f$$ totally differentiable at the origin?

My approach: find the partial derivatives over $$x$$ and $$y$$ in the origin. If they exist and are continuous, then $$f$$ is totally differentiable at the origin (answering b) and also we can calculate the directional derivative for any $$v$$ as $$\frac{df}{dv}(0,0) = f'(x)v$$

My first question: will this approach work? Secondly: I'm stuck calculating the partial derivatives: $$\lim_{h\rightarrow \infty} \frac{\sin(h^3)}{h^2}$$ How do I prove what that limit is?

You can compute $$\frac{\partial f}{\partial x}(0,0)$$ by the definition:$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h=\lim_{h\to0}\frac{\sin(h^3)}{h^3}=1$$and, by the same argument, $$\frac{\partial f}{\partial y}(0,0)=1$$. So, if $$f$$ is differentiable at $$(0,0)$$, you must have$$f'(0,0)(a,b)=\frac{\partial f}{\partial x}(0,0)a+\frac{\partial f}{\partial y}(0,0)b=a+b.$$
But $$f$$ is not differentiable at $$(0,0)$$, since this would mean that$$\lim_{(x,y)\to(0,0)}\frac{\bigl\lvert f(x,y)-f(0,0)-f'(0,0)(x,y)\bigr\rvert}{\lVert(x,y)\rVert}=0,$$which is equivalent to$$\lim_{(x,y)\to(0,0)}\frac{\bigl\lvert\sin(x^3+y^3)-x-y\bigr\rvert}{(x^2+y^2)^{3/2}}=0.$$However, this is not true; just see what happens if $$y=x$$.