[tex]\bf y=3-x^2\implies \cfrac{dy}{dx}=0-2x\implies \cfrac{dy}{dx}=-2x[/tex]
now, this is the equation to get the slope of the tangent line over that curve.
now, to get the normal, the normal is perpendicular to that tangent line, and thus it has a slope that is negative reciprocal to the tangent's.
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -2x\implies \cfrac{-2x}{1}\\\\
negative\implies +\cfrac{2x}{ 1}\qquad reciprocal\implies + \cfrac{ 1}{2x}\implies \left.\cfrac{1}{2x} \right|_{x=2}\implies \cfrac{1}{4}[/tex]
