Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.

Thank you, AmateurMathGuy and lhf.
Please check my work on Induction for Fibonacci sequence.
$$t_{k+2}-3t_{k+1}+t_{k}=0$$
$$y_{k+2}-3y_{k+1}+y_{k}=0$$
$(t_1,y_1)=(3,1)\to(x_1,y_1)=(1,1)$
$(t_2,y_2)=(7,3)\to(x_2,y_2)=(2,3)$
$(t_3,y_3)=(18,8)\to(x_3,y_3)=(5,8)$
$(t_4,y_4)=(47,21)\to(x_4,y_4)=(13,21)$
Since $1, 2, 3, 5, 8, 13, 21$ are in Fibonacci sequence, we predict that $(x_n,y_n)=(F_{2n-1},F_{2n})$ 
and will prove by Induction. 
$(x_n,y_n)=(F_{2n-1},F_{2n})$ is true for $n=1, 2, 3, 4$
Suppose that $(x_k,y_k)=(F_{2k-1},F_{2k})$ is true,
Since $y_{k+1}-3y_k+y_{k-1}=0$, so $y_{k+1}=3y_k-y_{k-1}=3F_{2k}-F_{2k-2}$
$=3F_{2k}-(F_{2k}-F_{2k-1})=2F_{2k}+F_{2k-1}=F_{2k}+F_{2k+1}=F_{2k+2}$
$x_{k+1}=\frac{t_{k+1}-y_{k+1}}{2}=\frac{(3t_k-t_k)-3y_{k-1}-y_{k-1}}{2}$
$=3\left(\frac{t_k-y_k}{2}\right)-\left(\frac{t_k-1-y_{k-1}}{2}\right)=3x_k-x_{k-1}$
Similarly, $x_{k+1}=F_{2k+1}$, so $(x_n,y_n)=(F_{2n-1},F_{2n})$
Answer : $(x_n,y_n)=(F_{2n-1},F_{2n})$

Cassini's identity  for the Fibonacci numbers is
$$
F_{n-1}F_{n+1}-F_{n}^{2}=(-1)^{n}
$$
Therefore
$$
1
=F_{2n-1}F_{2n+1}-F_{2n}^{2}
=F_{2n-1}(F_{2n-1}+F_{2n})-F_{2n}^{2}
=F_{2n-1}^2+F_{2n-1}F_{2n}-F_{2n}^{2}
$$
and $x=F_{2n-1}$, $y=F_{2n}$ are solutions of $x^2+xy-y^2 = 1$.
Similarly,
$$
F_{2n-1}^2-F_{2n-1}F_{2n-2}-F_{2n-2}^{2} = 1
$$
and $x=-F_{2n-1}$, $y=F_{2n-2}$ are solutions of $x^2+xy-y^2 = 1$. But the OP only wants positive solutions.
It remains to prove that these are the only solutions.
