How do i use chain rule to calculate the derivative dy/dx of a curve given by 2 "parametric equations": x=(t-1)^3, y=3t-8/t^2

(Explain why it is called "parametric"). The definition of chain rule says that we can re-write the derivative dy/dx in terms of the "parameter t" by dy/dx = (dy/dt) X (dt/dx). (Explain why this is the case).
So we have two equations, x and y, both equal to some perimeter t. To calculate the derivative dy/dx is simply the same as calculating the derivatives w.r.t "t" (either by inspection, product, or quotient rules) and then we multiply them together using the definition. That is dx/dt = 3(t-1)^2 , dy/dt = 3+16t^ (−3). So (using the definition) dy/dx=(3+16t^ (−3)) X ( 1/3(t-1)^2) = (3 + 16𝑡^−3)/(3(𝑡 − 1)^ 2).