Find dy/dx in terms of t for the curve defined by the parametric equations: x = (t-1)^3, y = 3t - 8/t^2, where t≠0

The first step is to recognise that, by the chain rule, dy/dx = dy/dt * dt/dx.
dy/dt and dt/dx can both be found by differentiating the functions given in the question, to give dy/dt and dx/dt. dt/dx is the inverse of dx/dt.
dy/dt = 3 + 16t^-3 by following the standard rule for differentiation ( y = x^n, dy/dx = nx^(n-1) )
dx/dt = 3(t-1)^2 by substitution. By saying the u = t-1, the function in y becomes x= u^3. Again using the chain rule, dx/dt = dx/du * du/dt. dx/du = 3u^2, and du/dt = 1. Therefore, dx/dt = 3(t-1)^2. Inverting this function gives dt/dx = 1/3*(t-1)^-2.
The final solution requires multiplying the functions for dy/dt and dt/dx to give dy/dx. dy/dx = dy/dt * dt/dx = (3 + 16t^-3) * 1/3*(t-1)^-2 = 3+16t^-3/3(t-1)^2