# Proving $(x^4-y^4) \cos (\frac{1}{\left\lVert (x,y) \right\rVert^3_2})$ is totally differentiable

How can one prove, that this function is totally differentiable on $$\mathbb{R^2}$$ and not continuous partially differentiable on $$\mathbb{R^2}$$?

$$f(x, y) := \begin{cases} (x^4-y^4)\cos\left(\dfrac{1}{\|(x,y)\|^3_2}\right), & (x, y) \neq (0,0); \\ 0, & (x, y) = (0, 0). \end{cases}$$

I know that to prove the total derivative one first has to check, whether this function is continuous and partially derivable, or not.

I also know that I can use the following formula to prove that a function is totally differentiable:

$$\dfrac{f(x, y) - f(0, 0) - \left(\left(\dfrac{\partial f}{\partial x}\right)(0, 0)\left(\dfrac{\partial f}{\partial y}\right)(0, 0)\right)\cdot\left({x-0}\atop{y-0}\right)}{|(x, y) - (0, 0)|}$$

If this formula gives a $$0$$, the function is totally differentiable.

I am stuck though, since I can't even find out if the function is continuous, or what the total derivative would be.

Can I substitute $$x^4 = a$$ and $$y^4 = b$$ and then we could follow:

$$(a-b)\cos\left(\frac{1}{\left\|\sqrt{\sqrt{(x,y)}} \right\|^3_2}\right) =$$

$$= (a-b)\cos\left(\frac{1}{\left\| \sqrt{(x,y)} \right\|^1_2}\right)$$

And then maybe l'Hospital (though I don't know how that would work for the denominator) and then calculating the limit for $$n\to\infty$$, which would be $$0\cdot 1$$ (I think, because $$\cos\left(\frac{1}{x}\right)\to 1$$ for $$\lim\to\infty$$), which gives us $$0$$, proving that the function is continuous.

First of the function is totally differentiable on $$\mathbb{R}^*$$

Next it is continuous around $$(0,0)$$ because $$|f(x,y)| < x^4+y^4$$

therefore $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$$

so f is continuous in $$(0,0)$$ and $$f(0,0) = 0$$

Next the gradient of $$f$$ is

$$\partial_xf = 4x^3\cos\left((x^2+y^2)^{-3/2}\right)-3(x^4-y^4)(x^2+y^2)^{-5/2}x\sin\left((x^2+y+2)^{-3/2}\right)$$

$$\partial_yf = -4y^3\cos\left((x^2+y^2)^{-3/2}\right)-3(x^4-y^4)(x^2+y^2)^{-5/2}y\sin\left((x^2+y+2)^{-3/2}\right)$$

so we expect that it is totally differentiable at $$(0,0)$$ and that $$\nabla f(0,0) = (0,0)$$

This is indeed true because

$$\left|\frac{f(x,y)}{(x^2+y^2)^{1/2}}\right| = \left|(x^2-y^2)(x^2+y^2)^{1/2}\cos\left((x^2+y^2)^{-3/2}\right)\right| \leq (x^2+y^2)^{3/2} \rightarrow_{(x,y)\rightarrow (0,0)} 0$$