Let $A\subseteq^*B\iff A$ \ $B$ is finite (For $A,B,\mathcal{P}(\Bbb N)$) i think i have might have written the question down incorrectly, i was wondering if someone can clarify should it be "(For $A,B\in \mathcal{P}(\Bbb N)$)" and how would i start to establish if $\subseteq^*$ is a partial ordering or a total ordering, would i go back to the definition of what $A$ \ $B$ is and go from there?
My attempt so far is the following: First recall that $A\setminus B=\{x\in A:x\notin B\}$, Now Let $B=\{x,y\}$ and let $A=\{x,y,z\}$ where $x,y,z \in \Bbb N$.Then clearly $A\setminus B=\{z\}$ which is finite and since $z\in \Bbb N$ then clearly $\{z\}\in \mathcal{P}(\Bbb N)$ then clearly $z\in A$ and so the reflexive property holds?. (i'm not sure my approach is correct any clarification of this would be great, i don't see any point in checking the other two properties are correct if my approach to the first property is incorrect).
When it says $A,B\in\mathcal{P}(\mathbb{N})$ it just means that $A$ and $B$ are subsets of $\mathbb{N}$.
To show that a relation is a partial order, you must show that it satisfies reflexivity, anti-symmetry, and transitivity. It sounds like you don't really understand what reflexivity means based on your attempt. A relation $R$ is reflexive iff $\forall x,xRx$. In your example, you need to show that for every set $A\in\mathcal{P}(\mathbb{N})$, $A\subseteq^* A$. This should be easy because $A\setminus A=\emptyset$, which is finite.
To show that a partial order relation $R$ is a total order, you must show that for all $A$ and $B$, either $ARB$ or $BRA$. To show that it isn't a total order, find a counterexample.
As was resolved in the comments below, $\subseteq^*$ isn't even a partial order because it doesn't satisfy antisymmetry. A counterexample is $\{1\} $ and $\{2\} $.