Find equations of the normal lines to the curve y=x^1/3 that is parallel to the line x+2y=1.
d/dx = 1/3x^2/3
d/dx = 1/3 x^-2/3
So, the slope of the normal line is -3x^(2/3)
x+2y has slope -1/2, so you need to find x such that
-3x^(2/3) = -1/2
x = (1/6)^(3/2)
y = 1/√6
So the equation of the normal is
y - 1/√6 = -1/2 (x - 1/(6√6))
I leave it to you to find the other line