Why is one relation transitive but the other is not? From what I have read about a transitive relation is that if xRy and yRz are both true then xRz has to be true. 
I'm doing some practice problems and I'm a little confused with identifying a transitive relation. 
My first example is a "equivalence relation"
$S=\{1,2,3\}$ and $R = \{(1,1),(1,3),(2,2),(2,3),(3,1),(3,2),(3,3)\}$
My Book solutions say that this relations is 
Reflexive and Symmetric
My Second example is "partial order"
$S=\{1,2,3\}$ and $R =\{(1,1),(2,3),(1,3)\}$
My Book solutions says is
Antisymmetric and Transitive
I got confused with why is the partial order(second example) Transitive. So what I did is that I applied $1R1$ and $1R3$ so $1R3$($xRy$ and $yRz$ so $xRz$). 
I tried to applied this to my first example (equivalence relation). What I did is $1R1$ and $1R3$ so $1R3$ ($xRy$ and $yRz$ so $xRz$).
Can someone explain what I'm missing or doing wrong? What can I do to identify a transitive relation? As you can see on both practice examples both have the same set of relations $1R1$ and $1R3$ so $1R3$($xRy$ and $yRz$ so $xRz$) but one is transitive and the other is not.

A relation $R$ of the set $S$ is transitive if:
$$\forall a{\in} S~\forall b{\in}S~\forall c{\in}S: \Big(\big((a,b){\in}R\wedge(b,c){\in}R\big)\to (a,c){\in}R\Big)$$
That definition is equivalent to:
$$\neg \exists a{\in}S~\exists b{\in}S~\exists c{\in}S: \Big(\big((a,b){\in}R\wedge(b,c){\in}R\big)\wedge (a,c){\notin}R\Big)$$
Thus what you are looking for are counter examples.   Look for any $(a,b), (b, c)$ in the relation without a matching $(a,c)$ in the relation.   Just one shows the relation is not transitive, but you have to be sure there are none to claim transitivity.

My first example is a "equivalence relation" $S=\{1,2,3\}$ and $R = \{(1,1),(1,3),(2,2),(2,3),(3,1),(3,2),(3,3)\}$ My Book solutions say that this relations is Reflexive and Symmetric

This is not transitive because while $(1,3)$ and $(3,2)$ are in the relation, $(1,2)$ is not.   One counter example is all we need.
PS: Because this is not transitive, it is not an equivalence relation.   Equivalence relations are those which are reflexive, symmetric, and transitive.

My Second example is "partial order" $S=\{1,2,3\}$ and $R =\{(1,1),(2,3),(1,3)\}$ My Book solutions says is Antisymmetric and Transitive

This is transitive because there is only one pair of $(a,b),(b,c)$ elements from which a counter example could be formed, $(\color{red}1,\color{blue}1),(\color{blue}1,\color{red}3)$, but $(\color{red}1,\color{red}3)$ is indeed in the relation; so there is no counter example.
