TXĐ: \(D = R\backslash \left\{ { - 2} \right\}\). Ta có: \(y' = \dfrac{4}{{{{\left( {x + 2} \right)}^2}}}\).

Phương trình tiếp tuyến của đồ thị hàm số tại điểm có hoành độ \(x = {x_0}\) là: \(y = \dfrac{4}{{{{\left( {{x_0} + 2} \right)}^2}}}\left( {x - {x_0}} \right) + \dfrac{{2{x_0}}}{{{x_0} + 2}}\,\,\left( d \right)\)

Gọi \(A = d \cap Ox\). Cho \(y = 0 \Rightarrow \dfrac{{4\left( {x - {x_0}} \right)}}{{{{\left( {{x_0} + 2} \right)}^2}}} + \dfrac{{2{x_0}}}{{{x_0} + 2}} = 0 \Leftrightarrow 4x - 4{x_0} + 2x_0^2 + 4{x_0} = 0 \Leftrightarrow x = \dfrac{{x_0^2}}{2} \Rightarrow A\left( {\dfrac{{x_0^2}}{2};0} \right)\)

\( \Rightarrow OA = \dfrac{{x_0^2}}{2}\) .

Gọi \(B = d \cap Oy\). Cho \(x = 0 \Rightarrow y = \dfrac{{ - 4{x_0}}}{{{{\left( {{x_0} + 2} \right)}^2}}} + \dfrac{{2{x_0}}}{{{x_0} + 2}} = \dfrac{{ - 4{x_0} + 2x_0^2 + 4{x_0}}}{{{{\left( {{x_0} + 2} \right)}^2}}} = \dfrac{{2x_0^2}}{{{{\left( {{x_0} + 2} \right)}^2}}} \Rightarrow B\left( {0;\dfrac{{2x_0^2}}{{{{\left( {{x_0} + 2} \right)}^2}}}} \right)\)  

\( \Rightarrow OB = \dfrac{{2x_0^2}}{{{{\left( {{x_0} + 2} \right)}^2}}}\)

\(\begin{array}{l} \Rightarrow {S_{\Delta OAB}} = \dfrac{1}{2}OA.OB = \dfrac{1}{2}.\dfrac{{x_0^2}}{2}.\dfrac{{2x_0^2}}{{{{\left( {{x_0} + 2} \right)}^2}}} = \dfrac{1}{{18}}\\ \Leftrightarrow 9x_0^4 = {\left( {{x_0} + 2} \right)^2} \Leftrightarrow \left[ \begin{array}{l}3x_0^2 = {x_0} + 2\\3x_0^2 =  - {x_0} - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0} = 1\\{x_0} = \dfrac{{ - 2}}{3}\end{array} \right.\end{array}\)

Vậy có 2 tiếp tuyến thỏa mãn là \(\left[ \begin{array}{l}y = \dfrac{4}{9}x + \dfrac{2}{9}\\y = \dfrac{9}{4}x + \dfrac{1}{2}\end{array} \right.\) .

Chọn A.