Solving Intersection Curve at (1,1,1): Derivatives & Tangent Line

[s3a]

Homework Statement

Given that near (1,1,1) the curve of intersection of the surfaces

x^4 + y^2 + z^6 - 3xyz = 0 and xy + yz + zx - 3z^8 = 0

has the parametric equations x = f(t), y = g(t), z = t with f, g differentiable.

(a) What are the values of the derivatives f'(1), g'(1)?
(b) What is the tangent line to the curve of intersection at (1,1,1) given that z = 1 + s? (Find what x and y are equal to.)

Answers:
f'(1) = 4
g'(1) = 7
x = 1 + 4s
y = 1 + 7s

Homework Equations

At least gradients, and partial derivative taking.

The Attempt at a Solution

I took the gradients of each surface (the first one being f(x,y,z) and the second one being g(x,y,z) respectively). I tried to set the respective components equal to each other in an attempt to solve for something but I ended nowhere useful. I also tried to think of z = t = 1 and in order to try and find x and y. Then, I was like "Oh, wait, I have a point (1,1,1), let me just shove it into the gradients." all in order to get the correct answers and then to try and make sense of what I did but I didn't succeed :(.

Any help in solving this problem would be greatly appreciated!
Thanks in advance!

 

[s3a]

Is what I wrote confusing or something? If it is, please tell me what to change that way I could get an answer.