Answer
tangent $y=-\frac{3}{2}x+\frac{5}{2}$
normal $y=\frac{2}{3}x+\frac{1}{3}$
Work Step by Step
Step 1. Given $x^3+y^2=2$, we have $3x^2dx+2ydy=0$ or $\frac{dy}{dx}=-\frac{3x^2}{2y}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(1,1)$, the slope of the tangent line is $m=-\frac{3}{2}$ and the equation is $y-1=-\frac{3}{2}(x-1)$ or $y=-\frac{3}{2}x+\frac{5}{2}$
Step 3. The slope of the normal line is $n=-1/m=\frac{2}{3}$ and the equation is $y-1=\frac{2}{3}(x-1)$ or $y=\frac{2}{3}x+\frac{1}{3}$
