Calculus Finding critical points

h(t) = t^(3/4) - 2t^(1/4)

3t^4 + 4t^3 - 6t^2

(y-1)/(y^2-y+1)

I understand how to find the derivative, but how do you finish it? Please explain in detail. Thanks

1.
Domain : t ≥ 0
h ' = 3/4 t^(3/4-1) - 2*(1/4)t^(1/4-1) = 0

3/4 t^(-1/4) - 1/2 t^(-3/4) = 0

3/4 t^(3/4) - 1/2 t^(1/4) = 0

t^(1/4) [ t^(2/4) - 1] = 0

t = 0 and t = 4/9 critical points

2.
3t^4 + 4t^3 - 6t^2
Domain: t ∈ R

f ' = 12 t^3 + 12 t^2 - 12 t = 0

t(t^2 + t - 1) = 0

t = 0
t = 1/2(-1 ± √(5))
critical points

3.
(y-1)/(y^2-y+1)

Domain y ∈ R
f ' = [y^2 - y + 1 - (y - 1)(2y - 1)] / (y^2-y+1)^2 = 0

y^2 - y + 1 -2y^2 + 3y - 1 = 0

y^2 - 2y =0

y(y - 2) = 0

critical points:
y = 0
y = 2