Answer
$f\left( {1,1} \right) = 2$
Work Step by Step
$$\eqalign{
& {\text{Maximize }}f\left( {x,y} \right) = \sqrt {6 - {x^2} - {y^2}} \cr
& {\text{Constraint: }}x + y - 2 = 0{\text{ }} \to {\text{ }}g\left( x \right) = x + y - 2 \cr
& {\text{Let }}f\left( {x,y} \right) = \sqrt {h\left( x \right)} ,{\text{ }}h\left( x \right) = 6 - {x^2} - {y^2} \cr
& {\text{then }}f\left( {x,y} \right){\text{ is maximum when }}h\left( x \right){\text{ is maximum}}{\text{.}} \cr
& \cr
& {\text{We need to maximize }}h\left( x \right) = 6 - {x^2} - {y^2}{\text{ }}\left( {{\text{objective function}}} \right) \cr
& {\text{Using the Method of Lagrange Multipliers}} \cr
& *{\text{ }}{h_x}\left( {x,y} \right) = \lambda {g_x}\left( {x,y} \right) \cr
& \frac{\partial }{{\partial x}}\left[ {6 - {x^2} - {y^2}} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {x + y - 2} \right] \cr
& - 2x = \lambda {\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& *{\text{ }}{h_y}\left( {x,y} \right) = \lambda {g_y}\left( {x,y} \right) \cr
& \frac{\partial }{{\partial y}}\left[ {6 - {x^2} - {y^2}} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {x + y - 2} \right] \cr
& - 2y = \lambda {\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{ and }}\left( {\bf{2}} \right) \cr
& x = y \cr
& \cr
& *{\text{ }}g\left( {x,y} \right) = c \cr
& x + y - 2 = 0 \cr
& x + y = 2 \cr
& {\text{Where }}x = y \cr
& x + x = 2 \cr
& x = 1 \to y = 1 \cr
& {\text{Therefore}}{\text{,}} \cr
& f\left( {1,1} \right) = \sqrt {6 - {{\left( 1 \right)}^2} - {{\left( 1 \right)}^2}} \cr
& f\left( {1,1} \right) = 2 \cr
& \cr
& {\text{The maximum is }}2{\text{ at }}x = 1,\,y = 1 \cr} $$