hcf(x,y) = 1 because if there are any solutions, there must be a smallest.
But since any common factor of x and y would divide both sides of the
equation in question i.e. 7(x+y)=3(x^2-xy+y^2), there wouldn't exist a
smallest
If we can show that hcf(x,y) =1, then from this it follows that:-
hcf(x+y, x^2-xy+y^2) = 3
Which gives only two options for x+y, = 3 or 9
and correspondingly for x^2 -xy+y^2, = 7 or 21
Of these (by trial and error) only the latter (9 & 21) works