Use the quotient rule and factor its numerator

For $u=2x^3+3$ and $v=x^4+2$, use $(u/v)^{\prime}=(vu^{\prime}-uv^{\prime})/v^2$. This gives $[(x^4+2)(6x^2)-(2x^3+3)(4x^3)]/(x^4+2)^2$.

As in the source annotation, factor $2x^2$ from the numerator. The remaining bracket is $(3x^4+6)-(4x^4+6x)=-x^4-6x+6$. Hence $y^{\prime}=2x^2(-x^4-6x+6)/(x^4+2)^2$. Since $x^4+2>0$, the formula is valid for every real $x$.