If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?

Suppose that the positive integer $d$ divides both $a+b$ and $a^2-ab+b^2$. You showed that $d$ divides $3ab$.
Note that $d$ is relatively prime to $ab$. Otherwise, there is a prime $p$ that divides both $d$ and $ab$. Since $p$ divides $ab$, it follows that $p$ divides one of $a$ or $b$, say $a$. But since $p$ divides $d$, it follows that $p$ divides $a+b$. Since $p$ divides $a$, it follows that $p$ divides $b$. This contradicts the fact that $a$ and $b$ are relatively prime.
Since $d$ divides $3ab$, and $d$ is relatively prime to $ab$, we conclude that $d$ divides $3$. So if $d$ is positive, then $d=1$ or $d=3$.
Note that both $1$ and $3$ are possible. For let $a=1$ and $b=1$. Then $\gcd(a+b,a^2-ab+b^2)=1$.
Now let $a=2$ and $b=1$. Then $a^2-ab+b^2=3$, so in this case the gcd is $3$.
