If cos y + 3 = xy3 - 4x then find dydx?

cos y + 3 = xy³ - 4x
(d/dx)(cos y + 3) = (d/dx)(xy³ - 4x)
-sin y•(dy/dx) = x•(3y²)•(dy/dx) + y³ - 4
(3xy² + sin y) (dy/dx) = 4 - y³
dy/dx = (4 - y³) / (3xy² + sin y)

If cos y + 3 = xy^3 - 4x 
y'(x) = (4 - y^3)/(3 x y^2 + sin(y))

cos(y) + 3 = xy^3 - 4x
- sin(y)dy/dx = [x* 3y^2dy/dx + y^3 * 1] - 4
- sin(y)dy/dx = 3xy^2dy/dx + y^3 - 4
- sin(y)dy/dx - 3xy^2dy/dx = y^3 - 4
dy/dx(- sin(y) - 3xy^2) = y^3 - 4
..............y^3 - 4
dy/dx = ----------------------
..............- sin(y) - 3xy^2

..............- (4 - y^3)
dy/dx=-----------------------
..............- [3xy^2 + sin(y)]

..............4 - y^3
dy/dx =---------------------- Answer//
.............3xy^2 + sin(y)

Notation: y' =dy/dx.;

cos(y) +3 = xy^3 -4x...(1).;
Implicit differentiation ---> -sin(y)*y' = y^3 -4 +(3xy^2)y';
[3xy^2 +sin(y)]y' = -(y^3-4);
(1)---> y^3-4 = (1/x)[cos(y) +3];
[3xy^2 + sin(y)]y' = -(1/x)[cos(y) +3];
y' = -(1/x)[cos(y) +3]/[3xy^2 +sin(y)] OR (4-y^3)/[3xy^2 +sin(y)].