find y' if x^2y^4+3y-4x^3=5 cos x-1

(2x-y^4)+(dy/dx4y^3*x^) +dy/dx+12x^2=5(-sin x-1)
is this started correctly

No, your approach is not correct.

To find y' (the derivative of y with respect to x) using implicit differentiation, we need to differentiate both sides of the equation with respect to x, treating y as an implicit function of x.

Let's break it down step by step:

1. Start with the given equation:
x^2y^4 + 3y - 4x^3 = 5cos(x) - 1

2. Differentiate both sides of the equation with respect to x using the chain rule.
The chain rule states that if we have a function f(g(x)), then its derivative with respect to x is f'(g(x)) * g'(x).

Differentiating the left side:
d/dx (x^2y^4) = 2x * y^4 + x^2 * d/dx (y^4)
Applying the chain rule, d/dx (y^4) = 4y^3 * dy/dx

Differentiating the middle term:
d/dx (3y) = 3 * dy/dx

Differentiating the right side:
d/dx (5cos(x) - 1) = -5sin(x)

3. Simplify the equation after differentiating:
2x * y^4 + x^2 * 4y^3 * dy/dx + 3 * dy/dx - 4 * 3x^2 = -5sin(x)

4. Rearrange the equation to solve for dy/dx:
Grouping like terms:
x^2 * 4y^3 * dy/dx + 3 * dy/dx = -2xy^4 - 12x^2 + 5sin(x)

Factoring out dy/dx:
(x^2 * 4y^3 + 3) * dy/dx = -2xy^4 - 12x^2 + 5sin(x)

Finally, solve for dy/dx (y'):
dy/dx = (-2xy^4 - 12x^2 + 5sin(x)) / (x^2 * 4y^3 + 3)

So the correct expression for y' is:
y' = (-2xy^4 - 12x^2 + 5sin(x)) / (x^2 * 4y^3 + 3)