Answer
The solution of the given equation is $\left\{ \frac{1}{3},\frac{-2}{3} \right\}$.
Work Step by Step
Consider the equation $9+\frac{3}{x}=\frac{2}{{{x}^{2}}}$.
Simplify:
$\begin{align}
& 9+\frac{3}{x}=\frac{2}{{{x}^{2}}} \\
& \frac{9{{x}^{2}}+3x}{{{x}^{2}}}=\frac{2}{{{x}^{2}}} \\
& 9{{x}^{2}}+3x-2=0
\end{align}$
Factorize and solve:
$\begin{align}
& 9{{x}^{2}}+3x-2=0 \\
& 9{{x}^{2}}+6x-3x-2=0 \\
& 3x\left( 3x+2 \right)-1\left( 3x+2 \right)=0 \\
& \left( 3x+2 \right)\left( 3x-1 \right)=0
\end{align}$
Thus,
$x=\frac{1}{3},\frac{-2}{3}$
Therefore, the required solution is $\left\{ \frac{1}{3},\frac{-2}{3} \right\}$.
