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
 A: 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 $y^3.$ 
A: $$(x^2-y^2)z-y^3=0$$
$$ x^2 z = y^2 (y + z) $$
Let $λ$ be the gcd of $x$ and $y$, then we have:
$$ x = λn, y=λm, (m, n) = 1$$ 
and the equation becomes:
$$ n^2 z = m^2 (λ m + z) $$
Then, because (m,n)=1, we must have: $z=m^2 ω$
$$ n^2 ω = (λ m + m^2 ω) $$
$$ n^2 ω = m (λ + m ω) $$
and again, because (m,n)=1, it must be: ω = m k
$$ n^2 k = (λ + m^2 k) $$
$$ (n^2 - m^2) k = λ $$
with the only restriction being: (m,n)=1
So, all the solutions are of the form:
$$ x = k n (n^2 - m^2) $$
  $$ y = k m (n^2 - m^2) $$
  $$ z = k m^3 $$ 
