the point is (3, 3 ^2/3).................. [³√3²=3 ^2/3]
Now for slope m,
dy/dx=d(x^2/3)/dx= 2/3 x^-1/3
at poin x=3
2* 3 ^-1/3/3= 2*(3 ^-1/3* 3 ^-1) = 2*3 ^-4/3

The line is perpendicular is m=-2^-1*3 ^4/3

finally the equation of the normal line is
y-3^2/3=-2^-1*3 ^4/3(x-3)

y = -2^-1*3 ^4/3 x + 2^-1*3^2/3 +2^-1*3 ^7/3

OR
y=(-3 ^4/3/2 )x +3 ^2/3 +3 ^7/3/2