Real analysis using field axioms to prove Prove that $(-x)(-y)=xy$ using field axiom theorems 

My attempt 

I know that $-(xy)$ is the additional inverse for $xy$, and I know $xy + -(xy) = 0$ (equation $1$)

I assumed that $(-x)(-y)=xy$ and substitute into eqn $1$ and I get

$(-x)(-y)+ -(xy) =0 $

so now I just have to show that this equation is still true to justify my claim that $xy=(-x)(-y)$

But how do I move on from here??

list of axioms : https://kevinbinz.files.wordpress.com/2014/11/group-theory-group-marriage-1.png
by associativity and commutativity of multiplication
we get 

$(-x)(-y) = (-1.x)(-1.y) = (-1.-1)(x.y)$

now remains to prove $-1.-1 = 1$

consider $-1 + -1.-1  = -1.1 + -1.-1 = -1(-1 + 1) = 0 $
by associativity and commutativity of multiplication
we get 

$(-x)(-y) = (-1.x)(-1.y) = (-1.-1)(x.y)$

now remains to prove $-1.-1 = 1$

consider $-1 + -1.-1  = -1.1 + -1.-1 = -1(-1 + 1) = 0 $