How can I find rational solutions for $x^3+y^3=1$ where both x and y are non-negative?
Edit: One of the answer in this post for general form of solutions
$$(a,b) \mapsto \left( \frac{a(a^3 + 2b^3)}{a^3 - b^3}, \; \; \frac{-b(2 a^3 + b^3)}{a^3 - b^3} \right).$$
It is obvious here that either $x$ or $y$ is negative.
If you had $$\left(\frac ab\right)^3 + \left(\frac cd\right)^3=1$$
then
$$(ad)^3 + (bc)^3=(bd)^3$$
which is a special case of $$p^3+q^3=r^3.$$
But it was proved by Euler no later than 1760 that this never happens, and possibly by Fermat as early as the 1630s.
The post you link to is about solutions of $x^3+y^3=9$, not of $x^3+y^3=1$. Obviously $1^3+2^3=9$. The situation is different because $1$ is a rational cube and $9$ isn't.