So, we have a third order linear homogeneous linear ODE and know two linearly independent solutions. And we need to find the third one using the method of the reduction of an order.

If we plug $y=zy_1$ we get for $z$ another third order linear homogeneous linear ODE which, however does not contain $z$ without derivatives, so denoting $z'=u$ we get a second order linear homogeneous linear ODE for $u$. But we know one of its linearly independent solutions, namely $u_1=(y_2/y_1)'$ and then plugging $u=vu_1$ we get another second order linear homogeneous linear ODE for $v$ which does not contain $v$ without derivatives so denoting $v'=w$ we get a first order linear homogeneous linear ODE for $u$.

However there is a little shortcut: since $u_1=1$ due to $y_1=t^2$ and $y_2=t^3$ the third order equation for $z$ contains neither $z$ nor $z'$ but only $z''$ and $z'''$ so denoting $z''=w$ we drop to the first order ODE just in one step instead of two.