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.