If \(\rm x\left(3-\frac{2}{x}\right)=\frac{3}{x}\) then the value of \(\rm x^2+\frac{1}{x^2}\) is

1
\(2\frac{1}{9}\)
2
\(2\frac{4}{9}\)
3
\(3\frac{1}{9}\)
4
\(3\frac{4}{9}\)

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