A cubic diophantine Equation While reading Diophantine equations I came across the following equation $$x^3+cy^3-3yx=0$$
Is there any known method to solve this equation for any $c$?

First note that the solutions involving zero are $x=y=0$ or $c=0,x=3t,y=3t^2$. Otherwise, we can give full solutions in terms of the (rare) solutions of much simpler equations. 

For $c\ne0$, the equation $x^3+cy^3=3xy$ is solvable in non-negative integers if and only if there is a solution in non-negative integers of the equation $$a^2u^3+bv^3=1\text{ or }3,\text{ where }c=ab.$$

Proof
Let $x=tz,y=tv$, where $z$ and $v$ are corime. Then $$t(z^3+cv^3)=3zv.$$
$z$ is a factor of $tcv^3$ and therefore of $tc$. Let $c=ab$ and $z=au$, where  $b$ and $u$ are coprime and where we can choose $b$ to be positive. Then $$t(a^2u^3+bv^3)=3uv.$$
Then $uv$ is coprime to $a^2u^3+bv^3$ and so $t=suv$ and  $$s(a^2u^3+bv^3)=3.$$
If this equation is solvable then the original equation has solution $x=asu^2v,y=suv^2.$ Note that switching the signs of $s,u$ and $v$ leaves $x$ and $y$ unchanged so we can suppose $s$ is positive.
Example 1  Solve $x^3+y^3=3xy.$
The equation $u^3+v^3=1$ has no non-zero solutions (Fermat) and the equation $u^3+v^3=3$ is impossible modulo $9$. So $x^3+y^3=3xy$ has no non-zero solutions. 
Example 2  Solve $x^3+6y^3=3xy.$
The equation $4u^3+3v^3=1 (s=3)$, for example, has solution $u=1,v=-1$, giving $x=-6,y=3.$
In general one can use PARI/GP to obtain all solutions. For example see the solution by @YongHaoNg to the following post:- 
Solutions of $ax^3+by^3=1$
