Find the equation of the tangent line to the curve 4x3+2y2−11=4xy−x at the point (−1,2).

Hello : let : y = f(x)     so : y' = f'(x)
calculate : f'(x)  and : f'(-1)  given f(-1) = 2

by derivate : 4x3+2y2−11=4xy−x
12x² +4yy' = 4y +4xy' - 1
4yy' - 4xy' = - 12x² +4y -1 
y' ( 4y - 4x) = 4y -12x² -1 ...(*)
if x = -1   y = 2 subsct in : (*)
y' (4(2)-4(-1)) = 4(2)-12(-1)² - 1
2y' = - 9 
y' = - 9/2 = f'(- 1)  ( the slope of the tangent )
of the tangent line to the curve at the point (−1,2) is : 
y - 2 = (-9/2)(x +1)