Do enunciado, podemos escrever: \begin{matrix} { f \circ g \ (x) = 3^{{ \left(x^2 + \frac{1}{x^2}\right) }} } &,& { h \circ f \ (x) = \dfrac{81}{ 3^{{ \left(x + \frac{1}{x}\right) }} } } \ \ \ \end{matrix}Assim, \begin{matrix} 3^{{ \left(x^2 + \frac{1}{x^2}\right) }} = \dfrac{81}{ 3^{{ \left(x + \frac{1}{x} \right)}}} \\ \\ 3^{ { \left(x^2 + \frac{1}{x^2} + x + \frac{1}{x} \right) } } = 3^4 \\ \\ \left(x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x}\right) = 4 \\ \\ \left(x^2 + \dfrac{1}{x^2}\right) + \left(x + \dfrac{1}{x}\right) = 4 \end{matrix}$\color{orangered}{Obs:}$ $\left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2$ \begin{matrix} \left[ \left(x + \dfrac{1}{x}\right)^2 - 2\right] + \left(x + \dfrac{1}{x}\right) = 4 \end{matrix}Seja $\left(x + \dfrac{1}{x}\right) = y$, \begin{matrix} \underbrace{y^2 + y - 6= 0} &\therefore& \fbox{$ y_1 = 2 \ \ , \ \ y_2 = -3$} \end{matrix}$• \ y_1$\begin{matrix} x + \dfrac{1}{x} = 2 \\ \\ \underbrace{x^2 - 2x + 1 = 0} \\ \fbox{$x_1=1$} \end{matrix} $• \ y_2$\begin{matrix} x + \dfrac{1}{x} = -3 \\ \\ \underbrace{x^2 + 3x + 1 = 0} \\ \\ \fbox{$x_2= \dfrac{-3 + \sqrt{5}}{2} \ \ , \ \ x_3= \dfrac{-3 - \sqrt{5}}{2}$} \end{matrix} Assim, o conjunto solução está $\color{orangered}{ \subset [-6,1]}$ \begin{matrix} Letra \ (C) \end{matrix}