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)
