Given function

#f(x)=\cot^2x-x^2#

Now, the slope of the tangent line #
frac{d}{dx}f(x)# is given as

#\frac{d}{dx}f(x)=\frac{d}{dx}(\cot^2x-x^2)#

#f'(x)=2\cot x(-\cosec^2 x)-2x#

#f'(x)=-2\cosec^2 x\cot x-2x#

Now, the slope of tangent at the point #x=\pi/3#

#f'(x=\pi/3)=-2\cosec^2(\pi/3)\cot(\pi/3)-2(\pi/3)#

#=-2(2/\sqrt3)^2(1/\sqrt3)-{2\pi}/3#

#=-8/{3\sqrt3}-{2\pi}/3#

Hence, the slope of normal at the same point #x=\pi/3# is

#=-1/(-8/{3\sqrt3}-{2\pi}/3)#

#={3\sqrt3}/{8+2\pi\sqrt3}#

Now, the y-coordinate of point #x=\pi/3# is given by substituting #x=\pi/3# in the given function,

#y=f(x=\pi/3)=\cot^2(\pi/3)-(\pi/3)^2#

#y=1/3-\pi^2/9#

Hence, the equation of normal to the given curve at #(\pi/3, 1/3-\pi^2/9)# & having slope #{3\sqrt3}/{8+2\pi\sqrt3}# is given as follows

#y-(1/3-\pi^2/9)={3\sqrt3}/{8+2\pi\sqrt3}(x-\pi/3)#

#y-1/3+\pi^2/9={3\sqrt3}/{8+2\pi\sqrt3}(x-\pi/3)#