Calculus Help?

Question

Consider the equation xz^2 − 6yz − 2 ln z = −6 as
defining z implicitly as a function of x and y. The
values of ∂z/∂x and ∂z/∂y at (0,1,1) are:

Answer 1

We are looking at the 3-D surface
f(x, y, z) = xz^2 - 6yz - 2 ln z + 6 = 0
∇f = (z^2, -6z, 2xz - 6y - 2/z) = (1, -6, -8) at (0, 1, 1).
We know if we make separate changes ∂x and ∂y to x and y, the change ∂z to z must be such that
(∂x, ∂y, ∂z) . (1, -6, -8) = 0, so that we stay on the surface.
For ∂z/∂x we let ∂y = 0 so we get ∂x = 8∂z and hence ∂z/∂x = 1/8. Similarly ∂z/∂y = -3/4.

Answer 2

dz/dx=(-z^2)/(2xz-6y-2/z) and dz/dy=(6z)/(-6y-2/z)

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