Problem: Given integers $a$ and $b$ prove that if there exist integers $x$ and $y$ for which $ax+by=(a,b)$ then $(x,y)=1.$ Note that $(a,b)=\gcd(a,b).$
Proof: Suppose $(x,y)=m,$ where $m\in\mathbb{Z}.$ Then $y=mk_1$ and $x=mk_2$ for some $k_1, k_2 \in \mathbb{Z}$. Let $(a,b)=d$. We now have $m((a/d)k_1+(b/d)k_2)=1.$ Since both numbers are integers we can conclude that $m=1$ which completes the proof.
Remark: I am unsure about the last claim and would therefore, be grateful if someone could either rectify my proof or declare that it is correct.
Your proof is correct. Here is an alternative. First of all, note that if $ax+by=c$ (where $a$, $b$, $c$, $x$ and $y$ are integers), then $(x,y)$ divides $c$ (since it has to divide any integer linear combination of $x$ and $y$).
Assume now that $(a,b)=1$ and $ax+by=1$. Then $(x,y)$ divides $1$, hence $(x,y)=1$. Now, if $d=(a,b)\neq 1$, then $\frac{a}{d}x+\frac{b}{d}y=1$ and $\frac{a}{d}$ and $\frac{b}{d}$ are integers, so $(x,y)=1$ by the previous case.