If x = -6 is a tangent to the curve x^3 + xy + 2y^3 = k, find all possible values of k

I calculated that:
dy/dx = (-3x^2 - y) / (6y^2 + x)
k≤(1/54)

f(x,y) = x³ + xy + 2y³ − k
∂f/∂y = xy' + 6y²y'
∂f/∂y = y' (x + 6y²)
∂f/∂y = 0

x + 6y² = 0
x = −6
−6 + 6y² = 0
y² − 1 = 0
y = ±1

k = −216 − 6 + 2 = −220
k = −216 + 6 − 2 = −212

x^3 + xy + 2y^3 = k
Differentiate w.r.to y
3x^2 dx/dy + x + y dx/dy + 6 y^2 = 0
3x^2 dx/dy + y dx/dy + = - 6 y^2 -x
dx/dy [3x^2 + y] = - 6 y^2 -x
dx/dy = (- 6 y^2 -x) / (3x^2 + y)
If x = -6 is a tangent, dx/dy = infinity
3x^2 + y = 0
y= - 3 x^2

k =x^3 + xy + 2y^3
= x^3 + x ( -3x^2) + 2 [ -3x^2] ^3
= x^3 - 3 x^3 - 54 x^6
= -2x^3 -54 x^6
= -2 x^3 ( 1 + 27x^3) when x=-6
= 2 * 216 ( 1- 27 * 216)
= - 2518992