Substituting #x=-3# in the given function: #f(x)=3^x-x# to get y-coordinate of given point as follows
#y=f(-3)#
#=3^{-3}-(-3)#
#=82/27#
Now, the slope of tangent #dy/dx# at any point to the given curve: #y=3^x-x# is given by differentiating given function w.r.t. #x# as follows
#dy/dx=f'(x)#
#=d/dx(3^x-x)#
#=3^x\ln3-1#
hence the slope of tangent at #(-3, 82/27)#
#f'(-3)=3^{-3}\ln3-1#
#={\ln3-27}/27#
hence the slope #(m)# of normal at the same point #(-3, 82/27)#
#m=-1/{f'(-3)}#
#=-1/{{\ln3-27}/27}#
#=27/{27-\ln3}#
hence the equation of normal at the point #(x_1, y_1)\equiv(-3, 82/27)# & having slope #m=27/{27-\ln3}# is given by following formula
#y-y_1=m(x-x_1)#
#y-82/27=27/{27-\ln3}(x-(-3))#
#y-82/27=27/{27-\ln3}(x+3)#
