How to show $\frac{1}{xy}$ = $\frac{1}{x}\frac{1}{y}$? I'm trying to prove that $\frac{1}{xy}$ = $\frac{1}{x}*\frac{1}{y},\forall x,y\neq 0$ using the field axioms of addition, multiplication, and the distributive law: $(x+y)z = xz+yz$ but am having a hard time doing so.
Things I've tried:
I've tried letting $\frac{1}{xy}$ = $\frac{1}{xy}$1 and $\frac{1}{xy}$=$\frac{1}{xy}$+0 and trying some fancy stuff, but so far I can't get anywhere with those.
Any help is greatly appreciated.

Remember the fields axioms (explained below). First you should think about which one to use in the statement of the desired identity. Certainly the demonstration through the axiom (M3). Second, you must realize that the secret of proof is to use the uniqueness of the axiom (M3). 
Fixed real numbers $a$, $b$ there are real numbers $\alpha$, $\beta$ and $\gamma$ such that $a\cdot \alpha =1$, $b\cdot \beta =1$ and $(b\cdot b)\cdot \gamma=1$. Let's $u,v$ and $w$ real numbers such that $a\cdot u =1$, $b\cdot v =1$ and $(a\cdot b)\cdot w=1$. The uniqueness assures us that $u=\alpha$, $v=\beta$ and $w=\gamma$. So if $\frac{1}{a}\cdot \frac{1}{b}=\alpha\cdot \beta$ satisfies identity ${a\cdot b}\left(\frac{1}{a}\cdot \frac{1}{b} \right)=1$, then we will have equality $\frac{1}{a\cdot b}=\left(\frac{1}{a}\cdot \frac{1}{b} \right)$. In fact,
\begin{align}
(x\cdot y)\cdot \left( \frac{1}{x}\cdot \frac{1}{y}\right)
&=
(x\cdot y)\cdot \left( \frac{1}{y}\cdot \frac{1}{x}\right)& \textrm{ by  (M4)} 
\\
&=
\left((x\cdot y)\cdot  \frac{1}{y}\right)\cdot \frac{1}{x}& \textrm{ by  (M1)} 
\\
&=
\left(x\cdot \left(y\cdot  \frac{1}{y}\right)\right)\cdot \frac{1}{x}&\textrm{ by  (M1) } 
\\
&=
\left(x\cdot 1\right)\cdot \frac{1}{x}& \textrm{ by  (M3)} 
\\
&=
x\cdot \frac{1}{x}& \textrm{ by  (M2)}
\\
&=
1&\textrm{ by  (M3)} 
\end{align}

  
*
  
*(M1) $(a\cdot b)\cdot c=a\cdot (b\cdot c), \quad \forall a,b,c\in \mathbb{R}-\{0\}$
  
*(M2) $\exists ! 1\in \mathbb{R}-\{0\} \textrm{ such that }  1\cdot a= a, \quad  \forall a\in\mathbb{R}-\{0\}$
  
*(M3) $\forall a\in \mathbb{R}-\{0\},\exists !\; \dfrac{1}{a}\in\mathbb{R} \textrm{ such that } a\cdot\dfrac{1}{a}=1$ 
  
*(M4) $a\cdot b= b\cdot a, \quad  \forall a,b\in \mathbb{R}-\{0\}$

