Answer
$$y - z = 0$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = \arctan \frac{y}{x},{\text{ }}\left( {1,0,0} \right) \cr
& z - \arctan \frac{y}{x} = 0 \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = z - \arctan \frac{y}{x} \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - \arctan \frac{y}{x}} \right] = - \frac{{ - \frac{y}{{{x^2}}}}}{{1 + {{\left( {y/x} \right)}^2}}} = \frac{y}{{{x^2} + {y^2}}} \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - \arctan \frac{y}{x}} \right] = - \frac{{\frac{1}{x}}}{{1 + {{\left( {y/x} \right)}^2}}} = - \frac{x}{{{x^2} + {y^2}}} \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - \arctan \frac{y}{x}} \right] = 1 \cr
& {\text{At the point }}\left( {1,0,0} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {1,0,0} \right) = \frac{0}{{{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}}} = 0 \cr
& {F_y}\left( {1,0,0} \right) = - \frac{1}{{{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}}} = - 1 \cr
& {F_z}\left( {1,0,0} \right) = 1 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& 0\left( {x - 1} \right) - 1\left( {y - 0} \right) + \left( {z - 0} \right) = 0 \cr
& {\text{Simplifying}} \cr
& - y + z + = 0 \cr
& y - z = 0 \cr} $$
