The normal at the point (1, 1) on the curve 2y = 3 - x2 is

$2 y=3-x^2 \Rightarrow 2 \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=-x$

⇒  slope of normal $=\frac{1}{x}$

∴  slope of normal at $(1,1)=\frac{1}{1}=1$

Equation of normal is $y-1=1(x-1) \Rightarrow x-y=0$