Answer
$f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = \sqrt 3 $
Work Step by Step
$$\eqalign{
& {\text{Maximize }}f\left( {x,y,z} \right) = x + y + z \cr
& {\text{Constraint: }}{x^2} + {y^2} + {z^2} = 1{\text{ }} \to {\text{ }}g\left( x,y,z \right) = {x^2} + {y^2} + {z^2} \cr
& \cr
& {\text{We need to maximize }}f\left( x,y,z \right) = x + y + z{\text{ }}\left( {{\text{objective function}}} \right) \cr
& {\text{Using the Method of Lagrange Multipliers}} \cr
& *{\text{ }}{f_x}\left( {x,y,z} \right) = \lambda {g_x}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial x}}\left[ {x + y + z} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2} + {z^2}} \right] \cr
& 1 = \lambda \left( {2x} \right) \cr
& 2\lambda x = 1{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& *{\text{ }}{f_y}\left( {x,y,z} \right) = \lambda {g_y}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial y}}\left[ {x + y + z} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2} + {z^2}} \right] \cr
& 1 = \lambda \left( {2y} \right) \cr
& 2\lambda y = 1{\text{ }}\left( {\bf{2}} \right) \cr
& \cr
& *{\text{ }}{f_z}\left( {x,y,z} \right) = \lambda {g_z}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial z}}\left[ {x + y + z} \right] = \lambda \frac{\partial }{{\partial z}}\left[ {{x^2} + {y^2} + {z^2}} \right] \cr
& 1 = \lambda \left( {2z} \right) \cr
& 2\lambda z = 1{\text{ }}\left( {\bf{3}} \right) \cr
& \cr
& {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{, }}\left( {\bf{2}} \right){\text{ and }}\left( {\bf{3}} \right) \cr
& {\text{Equating the equations }} \cr
& \frac{1}{{2x}} = \frac{1}{{2y}} = \frac{1}{{2z}} \cr
& x = y = z \cr
& \cr
& {\text{Substituting }}y = x{\text{ and }}z = x{\text{ into the constraint }} \cr
& {x^2} + {y^2} + {z^2} = 1 \cr
& {x^2} + {x^2} + {x^2} = 1 \cr
& 3{x^2} = 1 \cr
& x = \pm \frac{{\sqrt 3 }}{3} \cr
& {\text{Assuming that}}{\text{, }}x,{\text{ }}y,{\text{ and }}z{\text{ are positive}} \cr
& x = \frac{{\sqrt 3 }}{3},{\text{ }}y = \frac{{\sqrt 3 }}{3}{\text{ and }}z = \frac{{\sqrt 3 }}{3} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& f\left( {x,y,z} \right) = x + y + z \cr
& f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = \frac{{\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3} \cr
& f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = \sqrt 3 \cr
& \cr
& {\text{The maximum is }}\sqrt 3 {\text{ at }}x = y = z = \frac{{\sqrt 3 }}{3} \cr} $$