If $a\bmod{m}$ has an inverse x, and if $b\bmod{m}$ has an inverse y, prove that $ab\bmod{m}$ has an inverse xy? If $a\bmod{m}$ has an inverse x, and if $b\bmod{m}$ has an inverse y, prove that $ab\bmod{m}$ has an inverse xy?
Well knowing that $a\bmod{b}$ has an inverse when there exists an $i$ such that $ai\equiv 1\pmod{m}$
Here is what i think
$a\dot{}x=1(\mod{m})$ and $b\dot{}y=1(\mod{m})$ if we multiply both:
$a\dot{}x\dot{}b\dot{}y=1(\mod{m})$ from here on, 
$$ab\dot{}xy=1(\mod{m})$$
Would this actually prove it?

Yes, the proof is correct, assuming that you already know that such multiplication of congruences is valid, i.e. presuming known the Congruence Product Rule below. Recall also that inverses are necessarily unique: $ $ if $\,a,a'\,$ are inverses of $\,b\,$ then $\,a' = a'(ba) = (a'b)a = a.$ 

Below are proofs of common congruence rules. They state, essentially, that the equivalence relation of congruence is compatible with the (ring) operations '$+$' and '$*$'), i.e. the result of the operation does not depend upon which equivalence class representatives one chooses for its arguments.
Congruence Sum Rule $\rm\qquad\quad  A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#0a0}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#0a0}{A+B - (a+b)} $
Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \  B\equiv b\ \Rightarrow\ \color{#c00}{AB\equiv ab}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#c00}{AB - ab} $
Congruence Power Rule $\rm\quad\ A\equiv a\ \Rightarrow\ A^n\equiv a^n\ \  (mod\ m)$
Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, A^{n+1}\equiv a^{n+1},\,$ by the Product Rule, so the result follows by induction on $\,n.$
Beware $ $ that such rules need not hold true for other operations, e.g.
the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it reduces to the Power Rule, so follows by inductively applying $\,\rm b\,$ times the Product Rule).
