Find an equation of the tangent line to the curve at the given point.
y^2=x^3 (2-x) (1,1)
2yy' = (3x^2)(2-x) + (x^3)(-1)
= x^2(6-x-x)
= 2x^2(3-x)
y' = x^2(3-2x)/y
y'(1) = 1
So, now you have a point and a slope, so the line is
y-1 = 1(x-1)
y=x