I am confused as to how I derivate with multi-index.
I was trying to do my own example, let us choose $\alpha=(1,1)$ and my function $f(x,y)=x^3+y^3$. Then what is $\partial_{(x,y)}^{\alpha}f(x,y)$? By the notation in literature $\partial_{(x,y)}^{\alpha}f(x,y)= \partial_x^1\partial_y^1(x^3+y^3)$, then my first step is $\partial_y^1 (x^3+y^3)= 3y^2$ and after this $\partial_x^1 (3y^2)=0$?
Or would I need to calculate $\partial_x^1(x^3+y^3)\partial_y^1(x^3+y^3)$?
Sorry, I am not sure how this works. Please can somebody help me?
Your first interpretation is the accurate one; and it's fairly common. One way to think about it is that the operator $\partial^\alpha=\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}$ is in a sense the dual of the monomial $x^\alpha=x_1^{\alpha_1}\cdots x_d^{\alpha_d}$. Note that with this notation your function $f$ is $f(x)=x^{(3,0)}+x^{(0,3)}$ for $x=(x_1,x_2)$.
Note that more generally we have $\partial^\alpha x^\beta = \left( \partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\right) \left( x_1^{\beta_1}\cdots x_d^{\beta_d}\right)= \left( \partial_{x_1}^{\alpha_1}x_1^{\beta_1}\right)\cdots\left(\partial_{x_d}^{\alpha_d}x_d^{\beta_d}\right)$, which would vanish if there is at least one $i\in\{1,2,...,d\}$ such that $\alpha_i>\beta_i$.
One can visualize these partial derivatives e.g. in the case of two variables as the integer points in the first quadrant on the plane; see https://www.desmos.com/calculator/3gwhmhf6zf for a humble interactive graph. One can interpret the expression $\partial^\alpha x^\beta$ pictorially as follows: Consider the line passing through the origin and $\alpha$. Then for $\partial^\alpha x^\beta$, $\partial^\alpha$ takes the term at $\beta$ and moves it somewhere according to $-\alpha$; in particular the derivative vanishes if under this move $\alpha$ pushes $\beta$ out of the first quadrant.