for the equation given below, evaluate dy/dx at the point (-2,1). (3x-y)^4+2y^3=2403. dy/dx at (-2,1)=?

(3x - y)⁴ + 2y³ = 2403
(d/dx)[(3x - y)⁴ + 2y³] = (d/dx)(2403)
4(3x - y)³[3 - (dy/dx)] + 6y²(dy/dx) = 0
12(3x - y)³ - 4(3x - y)³(dy/dx)] + 6y²(dy/dx) = 0
-4(3x - y)³(dy/dx)] + 6y²(dy/dx) = -12(3x - y)³
2(3x - y)³(dy/dx)] + 3y²(dy/dx) = 6(3x - y)³
dy/dx = 6(3x - y)³ / [2(3x - y)³ - 3y²]

dy/dx at (-2, 1)
= 6[3(-2) - 1]³ / {2[3(-2) - 1]³ - 3(1)²}
= 6[-7]³ / {2[-7]³ - 3}
= -2058/-689
= 2058/689

(3x-y)^4 +2y^3 =2403
differentiate both sides with respect to x

4(3x-y)^3 (3-dy/dx) + 6y^2 dy/dx = 0
dy/dx (6y^2 - 4(3x-y)^3) = -12(3x-y)^3

dy/dx = -12(3x-y)^3 /(6y^2-4(3x-y)^3)
x=-2; y=1

dy/dx = 2058/689