Let us consider tangent to a curve #y=y(x)#, as shown below (in red line) at point #A#. To find the equation of tis tangent, we need the coordinates of the point #A#, as well as slope of tangent. This allows us to use point-slope form of the equation.
Consider the secant #AB#. If coordinates of #A# are #(x,y)#, our point #B# is given by #(x+Deltax,y+Deltay)#.
Observe that #y+Delta y=y(x+Deltax)#. The slope of the secant is given by #(Deltay)/(Deltax)#. Let us rotate this secant clockwise with #A# as pivot, which reduces distance between #AB# and#Deltax# too reduces, until #B# coincides with #A#, at which point #Deltax->0# and secant becomes the tangent.
Hence, the slope of tangent is #lim_(Deltax->0)(y(x+Deltax)-y(x))/(Deltax)#
or in other words slope of tangent is #(dy)/(dx)#
Here we have #f(x)=x^3#, therefore
#y(x+Deltax)=(x+Deltax)^3#
and slope of tangent is
#(dy)/(dx)=lim_(Deltax->0)((x+Deltax)^3-x^3)/(Deltax)#
= #lim_(Deltax->0)(x^3+3x^2Deltax+3x(Deltax)^2+(Deltax)^3-x^3)/(Deltax)#
= #lim_(Deltax->0)(3x^2Deltax+3x(Deltax)^2+(Deltax)^3)/(Deltax)#
= #lim_(Deltax->0)3x^2+3xDeltax+(Deltax)^2#
= #3x^2#
and slope at point #(1,1)# is #3xx1^2=3#. Now using point-slope form of equation, tangent at #(1,1)# is
#y-1=3(x-1)# or #y=3x-2#
graph{(y-x^3)(y-3x+2)=0 [-0.515, 1.985, -0.065, 1.185]}
