Setting ₩Gamma $f = (x, f(x))$ which is in $(X*Y)$ s.t. $x$ is an element of $X$) and $f:= X to Y$
We know this is a graph,
but how to show this is an equivalence relation
Reflexive $f(x)=f(x)$ so $x$~$x$
And symmetric, transitive conditions must be satisfied. But $f(x)=-x$ is bijection so there are no other element which satisfy $f(x)=f(y)$ if $x=y$
How to prove symmetricity and transitivity?
You're going wrong already for reflexitivy.
Your formatting is a bit wonky, but it looks like your definition is
$$ {\sim} = \{(x,f(x)) \mid x\in X\} $$
for some fixed function $f$. This means that $a\sim b$ if and only if there is some $x$ such that $a=x$ and $b=f(x)$. Or, cutting out the middleman, $a\sim b$ if and only if $b=f(a)$.
For reflexivity we then ask: Is it always the case that $x\sim x$? Or, by the definition we've just unfolded: Is it always the case that $x=f(x)$ -- no matter what $x$ is?
In your attempt you've somehow ended up writing $f(x)=f(x)$ instead of $x=f(x)$.
So when $f$ is negation, the relation is not reflexive. Namely, if $x=42$, then $x=f(x)$ does not hold: $42\ne -42$.
