Question

Use logarithmic differentiation to find y ′ for the following function.
x^3 = e^−4y4−9y

Answer 1

I assume you mean

x^3 = e^(−4y^4−9y)

otherwise using logs makes no sense. So, we have

3logx = -4y^4-9y
3/x dx/dy = -16y^3-9
dx/dy = -x/3 (16y^3+9)
so, dy/dx = -3/[x(16y^3+9)]

Not the usual formulation, eh?

using implicit differentiation, we get

3x^2 = -(16y^3+9)e^(−4y^4−9y) y'
y' = -3x^2/[e^(−4y^4−9y)(16y^3+9)]
But that bottom factor of e^(−4y^4−9y) is just x^3, so you again have
y' = -3/[x(16y^3+9)]