Answer
The tangent at $(1,1)$ is $$y=2x-1$$
The normal at $(1,1)$ is $$y=-\frac{1}{2}x+\frac{3}{2}$$
Work Step by Step
$$y^2(2-x)=x^3$$
1) Find the derivative of the function using implicit differentiation:
$$(y^2)'(2-x)+y^2(2-x)'=3x^2$$ $$2yy'(2-x)+y^2(-1)=3x^2$$ $$2yy'(2-x)-y^2=3x^2$$ $$2yy'(2-x)=3x^2+y^2$$ $$y'=\frac{3x^2+y^2}{2y(2-x)}$$
2) The slope of the tangent line to the curve at $(1,1)$ is $$y'=\frac{3\times1^2+1^2}{2\times1(2-1)}=\frac{4}{2\times1}=2$$
The tangent line to the curve at $(1,1)$, therefore, is $$y-1=2(x-1)$$ $$y-1=2x-2$$ $$y=2x-1$$
3) As tangent line is perpendicular with its corresponding normal line, the product of their slopes equal $-1$.
In other words, if we call the slope of the normal at $(1,1)$ $k$, then we have $$k\times2=-1$$ $$k=-\frac{1}{2}$$
The normal line to the curve at $(1,1)$, therefore, is $$y-1=-\frac{1}{2}(x-1)$$ $$y-1=-\frac{1}{2}x+\frac{1}{2}$$ $$y=-\frac{1}{2}x+\frac{3}{2}$$
