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?

$\gcd(a+b, a^2-ab+b^2) = \gcd(a+b, (a+b)^2-3ab) = \gcd(a+b, 3ab)$ by the Euclidean algorithm. If the gcd was $d \ne 1 ,3$, then $d \mid a$ or $d \mid b$ in $3ab$ but then from $a+b$, $d$ would divide the other. Thus, the result follows. 
