Answer
$y = 1.5x - 0.5$
$y = -\frac{2}{3}x + 1\frac{2}{3}$
Work Step by Step
first of all lets find the derivative of y:
$y^{2} = x^{3}$
$y = \sqrt (x^{3})=x^{1.5}$
$y' = (x^{1.5})'= 1.5x^{0.5}$
So the slope of the tangent line at $(1, 1)$ is (x = 1, and calculate the derivative)
$1.5\times 1^{0.5}=1.5$
We use the point-slope form to write an equation of the tangent line at $(1,1)$
$y - 1 = 1.5 (x-1)$
$y = 1.5x - 1.5 + 1 = 1.5x - 0.5$
The slope of the normal line at $(1,1)$ is the negative reciprocal of $1.5$, namely $-\frac{2}{3}$, so an equation is
$y - 1 = -\frac{2}{3}(x-1)$
$y = -\frac{2}{3}x +\frac{2}{3} + 1 = -\frac{2}{3}x + 1\frac{2}{3}$