Find all solutions in N of the following Diophantine equation $(x^2 − y^2)z − y^3 = 0$
i divide by $z^3$ and look for rational solutions of the equation $A^2 − B^2 − B^3 = 0.$  The point $(A,B) = (0, 0)$ is a singular point, that is any line
through this point will meet the curve twice in $(0, 0)$. Now i wanna use Diophantus chord method using the lines
passing through $(0, 0)$ but i can't seem to pass this point

At least for a nonvertical line $B=kA$ when plugged into 
$$A^2-B^2-B^3=0 \tag{1}$$ gives
$$A^2-k^2A^2-k^3A^3=0,$$
where here the double root corresponds to $A^2$ being a factor. After dividing by that and solving for $A$ one gets $A=(1-k^2)/k^3,$ and so also $B=kA=(1-k^2)/k^2.$ Finally a check reveals these values of $A,B$ satisfy $(1).$ So this is a parametrization of the solution set for rationals, which covers most points on the curve.
There of course would still be a lot of work to go from such a rational parametrization to finding all positive integer solutions to the starting equation.
Added note: For the starting equation $$(x^2-y^2)z-y^3=0, \tag{2}$$ if one puts the $k$ parameter in the above rational parametrization equal to $m/n,$ where we assume $0<m<n$ in order that we have $0<k<1,$ then we can work backwards and get a collection of positive integer solutions $(x,y,z)$ to equation $(2).$ The resulting expressions are
$$x=n(n^2-m^2),\ \ y=m(n^2-m^2),\ \ z=m^3.\tag{3}$$
Note that if given a positive integer solution to $(2)$ we multiply each of $x,y,z$ by some positive integer $k$ say, another solution results. So one could call a solution "primitive" provided the gcd of $x,y,z$ is $1$. In this sense, the remaining question is whether there are primitive solutions not included in the equations $(3).$ I've had some success so far on this, having shown at least that $z$ must be a cubic factor of $y^3.$ 
