What is the slope of the tangent line through (3,2) and (x, y) for x=-2 and y=x3

First you must derive the equation to find the slope:
  y = x ^ 3
 y '= 3x ^ 2
 Then, for x = -2 we have the slope is:
 y '(- 2) = 3 (-2) ^ 2 = 12
 Then the equation of the line with slope equal to 12 and passing through point (3,2) is:
 y = 12 (x-3) +2

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