Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that
The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.
Alternatively, it's a quadratic in $x$, with solution
for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)