As explained by J.C. Ottem, the curves of degree $6$ are contained in a cubic. Let us assume that the surface is smooth, so you can see the cubic surface as the blow-up of six points in $\mathbb{P}^2$ with no $3$ collinear and not all on the same conic.

The surface $X$ has Picard group generated by $L$, the preimage of a line in $\mathbb{P}^2$, and by $E_1,\dots,E_6$, the exceptional divisors.

The trace of an hyperplane is the anti-canonical divisor, $-K_X=3L-E_1-\dots-E_6$.

$1)$ To obtain degree $6$ rational curves in $X$, you take a conic of $\mathbb{P}^2$ not passing through the points. It is equivalent to $C=2L$. You can thus see that no quadric of $\mathbb{P}^3$ contains it, because $C$ it is not equivalent to $-2K_X$. There is a cubic $X'$ which contains $C$ is and only if $-3K_X-C=7L-3E_1-\dots-3E_6$ is effective. Intersecting this system with the line $2L-E_1-\dots-E_6+E_i$, we obtain $-1$, so the system should contain the $6$ lines, which is not possible. Hence, the cubic $X$ is the unique one which contains $C$. However, there are quartics containing $C$, because $-4K_X-C=10L-4E_1-\dots-4E_6$ is effective. This can be seen by taking the union of two general quintics of $\mathbb{P}^2$ with $6$ double points. These curves are other rational sextics of $\mathbb{P}^3$. The system $-4K_X-C$ being moreover base-point free, the ideal of $C$ in $\mathbb{P}^3$ is generated by the cubic and equations of degree $4$. You can do this easily in coordinates for a given curve, but this takes time.

$2)$ To obtain degree $6$ curves of genus $1$, you can take a cubic of $\mathbb{P}^2$ passing through three of the points, i.e. $C=3L-E_1-E_2-E_3$. The same type of calculation gives you the ideal.

You can of course continue with all curves of degree $6$. If you want more details on the type of the curves you obtain on the cubic surface, see "Weak Fano threefolds obtained by blowing-up a space curve and construction of Sarkisov links" by S. Lamy and myself, especially Proposition 4.2.