You have quoted the result essentially correctly. We want the tangent line at the point on the curve $y=f(x)$ which has $x$-coordinate say $x_1$. Calculate the $y$ coordinate $y_1$, using $y_1=f(x_1)$. Then the equation of the tangent line can be written as
In our situation, we have $f(x)=Ax^3+Bx^2+Cx+D$, so as you wrote we have $f'(x)=3Ax^2+2Bx+C$. Thus
Remark: The number $f'(x_1)$ gives the slope of the tangent line at the point on the curve which has $x$-coordinate equal to $x_1$. Now if you take a "general" point $(x,y)$ on the tangent line, then the slope of the line that joins this point $(x,y)$ to $(x_1,y_1)$ is equal to
(the change in $y$ divided by the change in $x$). Since this slope is supposed to be $f'(x_1)$, we get
Now multiply both sides by $x-x_1$ to get the equation of the tangent line that you quoted.