Is the empty set partially ordered ? Also, is it totally ordered?

The question as phrased isn’t really meaningful, since you didn’t specify a relation on the empty set. However, there is only one, so I’ll assume that it’s the one that you meant. A relation on a set $A$ is a subset of $A\times A$. Since $\varnothing\times\varnothing=\varnothing$, the only subset of $\varnothing\times\varnothing$ is $\varnothing$. Thus, the question can be interpreted as:

Is $\varnothing$ a partial order on $\varnothing$? Is it a total order?

Let’s recall the definitions

A relation $R$ on a set $A$ is a partial order if it is reflexive, antisymmetric, and transitive.

• Reflexive: For each $a\in A$, $a\,R\,a$.

• Antisymmetric: For all $a,b\in A$, if $a\,R\,b$ and $b\,R\,a$, then $a=b$.
• Transitive: For all $a,b,c\in A$, if $a\,R\,b$ and $b\,R\,c$, then $a\,R\,c$.

If in addition it’s true that for all $a,b\in A$, $a\,R\,b$ or $b\,R\,a$, then $R$ is a total order on $A$.

The thing to notice here is that each of these is a universally quantified statement: something must be true of every element, pair of elements, or trio of elements of $A$. Thus, in order to show that $R$ is not reflexive, you must find an $a\in A$ such that $a\,\not R\,a$; in order to show that $R$ is not antisymmetic, you must find elements $a,b\in A$ such that $a\,R\,b$ and $b\,R\,a$, but $a\ne b$; in order to show that $R$ is not transitive, you must find elements $a,b,c\in A$ such that $a\,R\,b,b\,R\,c$, and $a\,\not R\,c$; and in order to show that $R$ is not total, you must find elements $a,b\in A$ such that $a\,\not R\,b$ and $b\,\not R\,a$. The crucial point is that in each case you must find elements of $A$ that actually have certain properties. If such elements don’t exist, then $R$ is reflexive (or antisymmetric, transitive, or total).

In your question $A$ is $\varnothing$, the empty set. Is it possible to find elements of $\varnothing$ that have certain properties? Is it possible to find elements of $\varnothing$ at all?

The above answers confirm that $\emptyset$ with the empty ordering is in fact a linear order. I would just like to add that it is in fact also well-ordered under this ordering since it is linearly ordered and every non-empty subset of $\emptyset$ has a least element. This is true vacuously since there are no nonempty subsets at all.

If a set is totally ordered, then it is already partially ordered.

It is meaningless to say whether any set $A$ is partially or totally ordered. You need to give a set $A$ and a partial ordering $\prec$ on $A$ for this question to make sense.

However, there does exists a total ordering on the $\emptyset$. It is the empty relation. More concretely, you can also think of the $\in$ relation restricted to $\emptyset$.