Answer
$$\theta \approx 25.23^\circ $$
Work Step by Step
$$\eqalign{
& 2xy - {z^3} = 0,{\text{ }}\left( {2,2,2} \right) \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = 2xy - {z^3} \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 2y \cr
& {F_y}\left( {x,y,z} \right) = 2x \cr
& {F_z}\left( {x,y,z} \right) = - 3{z^2} \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr
& \nabla F\left( {x,y,z} \right) = {F_x}\left( {x,y,z} \right){\bf{i}} + {F_y}\left( {x,y,z} \right){\bf{j}} + {F_z}\left( {x,y,z} \right){\bf{k}} \cr
& \nabla F\left( {x,y,z} \right) = 2y{\bf{i}} + 2x{\bf{j}} - 3{z^2}{\bf{k}} \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right){\text{ at the point }}\left( {2,2,2} \right) \cr
& \nabla F\left( {2,2,2} \right) = 2\left( 2 \right){\bf{i}} + 2\left( 2 \right){\bf{j}} - 3{\left( 2 \right)^2}{\bf{k}} \cr
& \nabla F\left( {2,2,2} \right) = 4{\bf{i}} + 4{\bf{j}} - 12{\bf{k}} \cr
& {\text{The angle of inclination of the tangent plane is}} \cr
& \cos \theta = \frac{{\left| {\nabla F\left( {2,2,2} \right) \cdot {\bf{k}}} \right|}}{{\left\| {\nabla F\left( {2,2,2} \right)} \right\|}} \cr
& \cos \theta = \frac{{\left| {\left( {4{\bf{i}} + 4{\bf{j}} - 12{\bf{k}}} \right) \cdot {\bf{k}}} \right|}}{{\left\| {4{\bf{i}} + 4{\bf{j}} - 12{\bf{k}}} \right\|}} \cr
& \cos \theta = \frac{{\left| { - 12} \right|}}{{\sqrt {16 + 16 + 144} }} \cr
& \cos \theta = \frac{{12}}{{\sqrt {176} }} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{{12}}{{\sqrt {176} }}} \right) \cr
& \theta \approx 25.23^\circ \cr} $$
