Preliminaries: Just to be clear, when I use, e.g. $x \mid y$, that means $x$ divides $y$, (or, alternatively, that $y$ is divisible by $x$
Let's look at symmetry.
Let's pick even integers $x = 2, y=4$, just to test one case.
Then $x\mid y,$ (because $2$ divides $4$, since $2\cdot 2 = 4$), but $y$ does not divide $x$ (because there is no positive integer $k$ such that $4k = 2$).
All we need is one counterexample to prove that the relation is not symmetric, because a symmetric relation requires that for all $x, y,\;\;$ if $\;x\mid y,\;\;$ then $\;\;y\mid x$.
We see that doesn't hold $x = 2, y = 4.$ Therefore, it doesn't hold for all $x, y$ such that $x\mid y$. Hence the relation, as noted, is not symmetric.
Let's look at antisymmetry.
Now there are some cases in which $x$ divides $y$, and also $y$ divides $x$. When does that happen?
This relation is antisymmetric if, for all $x, y$, whenever it happens that $x$ divides $y$ AND also $y$ divides $x$, then it must be the case that $x = y$.
So let's suppose it happens that $x$ divides $y$, and $y$ divides $x$.
Then by definition, $y=xn$ and $x = ym$, where $n, m$ are positive integers.
We can substitute $\;y = xn\;$ into the equation $x= ym = (xn)m = nmx.$ Clearly, if $x = xnm,$ then $nm = 1$. And the only way that two positive integers, when multiplied, can equal one is if they are both equal to $1$, i.e. $n=m=1$. In short, we have that $y=x$, and $x=y$. Hence, when $x\mid y$ and $y\mid x$, it follows that $x=y$.
Hence the relation D is, in fact, antisymmetric.