If $P \in \mathbb{R}[x]$ is a polynomial, under which sufficient condition is the function: $$f: \mathbb{R}^2 \to \mathbb{R}: f(x,y) = P(|x+y|)$$ total differentiable?
So for a function to be total differentiable, every partial derivative need to be continuous or the function needs to fulfill the definition of total differentiability: $$\lim_{x \to a} \frac{||f(x) - f(a) -A(x-a)||}{||x-a||} = 0$$ where A is the linear map with matrix $A = (df)(a).$
So the first condition I have is that $x+y \neq 0 $. But I have the feeling that there needs be a more general or specific condition.
Clearly $f$ is differentiable at points where $x+y\ne 0$ since $|\cdot|$ is differentiable on $\mathbb{R}\setminus \{0\}$.
If we write $P(t) = \sum_{i=0}^na _i t^i$ then $f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+\sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|\cdot|^\alpha$ is differentiable for $\alpha > 1$.
Conversely, if $f(x,y) = \sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - \underbrace{\left( a_0+\sum_{i=2}^n a_i|x+y|^i\right)}_{\text{this is differentiable}} = a_1|x+y|$$ is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.