find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
then find the equations of the tangent and the normal to the curve at the point where x = 3
thanks
$\displaystyle f(x) = x^3 - 3x^2$
$\displaystyle f'(x) = 3x^2 - 6x$
$\displaystyle f'(3) = 3(3)^2 - 6(3) = 9$
So the slope of your tangent is $\displaystyle 9$, and the slope of your normal is $\displaystyle -\frac{1}{9}$.
Also, since
$\displaystyle f(3) = (3)^3 - 3(3)^2 = 0$,
both equations will pass through the point $\displaystyle (3, 0)$.
Let us know if you need more help from there.