If every pair satisfies $aRb\rightarrow bRa$ then the relation is symmetric. If there is at least one pair which fails to satisfy that then it is not symmetric.
Similarly if there is at least one pair which has $(aRb\rightarrow bRa)\land a\neq b$ then antisymmetry is also not satisfied.
We can therefore take the following relation: $\{a,b,c\}$ would be our universe and $R=\{\langle a,b\rangle,\langle b,a\rangle,\langle a,c\rangle\}$.
The fact that $aRc\land\lnot cRa$ shows that the relation is not symmetric, but $a\neq b$ and both $aRb$ and $bRa$ hold.
Here’s a way to think about symmetry and antisymmetry that some people find helpful. A relation $R$ on a set $A$ has a directed graph (or digraph) $G_R$: the vertices of $G_R$ are the elements of $A$, and for any $a,b\in A$ there is an edge in $G_R$ from $a$ to $b$ if and only if $\langle a,b\rangle\in R$. Think of the edges of $G_R$ as streets. The properties of symmetry, antisymmetry, and reflexivity have very simple interpretations in these terms:
$R$ is reflexive if and only if there is a loop at every vertex. (A loop is an edge from some vertex to itself.)
$R$ is symmetric if and only if every edge in $G_R$ is a two-way street or a loop. Equivalently, $G_R$ has no one-way streets between distinct vertices.
$R$ is antisymmetric if and only every edge of $G_R$ is either a one-way street or a loop. Equivalently, $G_R$ has no two-way streets between distinct vertices.
This makes it clear that if $G_R$ has only loops, $R$ is both symmetric and antisymmetric: $R$ is symmetric because $G_R$ has no one-way streets between distinct vertices, and $R$ is antisymmetric because $G_R$ has no two-way streets between distinct vertices.
To make a relation that is neither symmetric nor antisymmetric, just find a digraph that has both a one-way street and a two-way street, like this one:
$$0\longrightarrow 1\longleftrightarrow 2$$
It corresponds to the relation $R=\{\langle 0,1\rangle,\langle 1,2\rangle,\langle 2,1\rangle\}$. on $A=\{0,1,2\}$.
Best Answer
A convenient way of thinking about these properties is from a graph-theoretical perspective.
Let us define a graph (technically a directed multigraph with no parallel edges) in the following way:
Have a vertex for every element of the set. Draw an edge with an arrow from a vertex $a$ to a vertex $b$ iff there $a$ is related to $b$ (i.e. $aRb$, or equivalently $(a,b)\in R$).
If an element is related to itself, draw a loop, and if $a$ is related to $b$ and $b$ is related to $a$, instead of drawing a parallel edge, reuse the previous edge and just make the arrow double sided ($\leftrightarrow$)
For example, for the set $\{1,2,3\}$ the relation $R=\{(1,1), (1,2), (2,3), (3,2)\}$ has the following graph:
Definitions:
$\begin{array}{l|l|l} &\text{set theoretical}&\text{graph theoretical}\\ \hline \text{Symmetric}&\text{If}~aRb~\text{then}~bRa&\text{All arrows (not loops) are double sided}\\ \hline \text{Anti-Symmetric}&\text{If}~aRb~\text{and}~bRa~\text{then}~a=b&\text{All arrows (not loops) are single sided} \end{array}$
You see then that if there are any edges (not loops) they cannot simultaneously be double-sided and single-sided, but loops don't matter for either definition. Any relation on a set $A$ that is both anti-symmetric and symmetric then has its graph consisting of only loops (i.e. is of the form $R=\{(a,a)~|~a\in S\subseteq A\}$ for some $S\subseteq A$.
Any relation whose graph contains both types of arrows (single-sided and doublesided) will be neither symmetric nor antisymmetric.