f(x) =sqrt(x^6) = x^3
f'(x) = 3x^2
Also, 6x - 2y +10 =0
or 2y = 6x +10
or y = 3 x +5,
Since, tangent from f(x) is parallel to the line 6x - 2y +10 =0, so
3x^2 = 3
x^2 = 1
x = +/- 1
x = 1 or -1
then f(1) = 1 and f(-1) = -1
So the equation of line will be
y-1 = 3(x-1) if f(1) or
y+1 = 3(x+1) if f(-1)