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)#