Jtdylktuy

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

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

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:

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

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:

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

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

And then maybe l'Hospital (though I don't know how that would work for the denominator) and then calculating the limit for $ntoinfty$, which would be $0cdot 1$ (I think, because $cosleft(frac{1}{x}right)to 1$ for $limtoinfty$), 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

$$partial_xf = 4x^3cosleft((x^2+y^2)^{-3/2}right)-3(x^4-y^4)(x^2+y^2)^{-5/2}xsinleft((x^2+y+2)^{-3/2}right)$$

$$partial_yf = -4y^3cosleft((x^2+y^2)^{-3/2}right)-3(x^4-y^4)(x^2+y^2)^{-5/2}ysinleft((x^2+y+2)^{-3/2}right)$$

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

This is indeed true because

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