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Rational inequality is given as$\frac{1}{3}+\frac{1}{{\mathrm{x}}^2}>\frac{4}{3\mathrm{x}}.$

On subtracting$\frac{4}{3\mathrm{x}}$on both sides,

$$\begin{gathered} \frac{1}{3}+\frac{1}{{\mathrm{x}}^2}>\frac{4}{3\mathrm{x}} \\ \frac{1}{3}+\frac{1}{{\mathrm{x}}^2}-\frac{4}{3\mathrm{x}}>\frac{4}{3\mathrm{x}}-\frac{4}{3\mathrm{x}} \\ \frac{{\mathrm{x}}^2+3-4\mathrm{x}}{3{\mathrm{x}}^2}>0 \end{gathered}$$

For the definition of the above integral, the denominator should be positive not even zero. Thus, one of the critical points is$\mathrm{x}=0.$

To true this inequality, the condition can be denoted as,

$$\begin{gathered} {\mathrm{x}}^2+3-4\mathrm{x}>0 \\ {\mathrm{x}}^2-4\mathrm{x}+3>0 \end{gathered}$$

The polynomial function is denoted as,

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-4\mathrm{x}+3$

The polynomial function can be factorized using AC method.

The form of quadratic equation is,

$$\begin{gathered} {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ {\mathrm{x}}^2-4\mathrm{x}+3 \\ {\mathrm{x}}^2-4\mathrm{x}+3>0 \\ {\mathrm{x}}^2-3\mathrm{x}-\mathrm{x}+3>0 \end{gathered}$$

On solving and factorizing,

$$\begin{gathered} \mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)>0 \\ (\mathrm{x}-3)(\mathrm{x}-1)>0 \end{gathered}$$

To find the critical point,

The critical points are$\mathrm{x}=0,1,3.$

The value of $\frac{{\mathrm{x}}^2+3-4\mathrm{x}}{3{\mathrm{x}}^2}$can be tested with points,

$$\begin{gathered} \mathrm{x}=-1\mathrm{is}\frac{{(-1)}^2+3+4}{3\left(1\right)}=\frac{8}{3} \\ \mathrm{x}=0.5\mathrm{is}\frac{{(0.5)}^2+3-(4×0.5)}{3{(0.5)}^2}=\frac{5}{3} \\ \mathrm{x}=2\mathrm{is}\frac{{\left(2\right)}^2+3-8}{3\left(4\right)}=\frac{-1}{12} \\ \mathrm{x}=5\mathrm{is}\frac{{\left(5\right)}^2+3-20}{3\left(5\right)}=\frac{8}{75} \end{gathered}$$

To true this inequality, the quotient should be positive not even zero.

This satisfies$\mathrm{x}∈(-∞,0)∪(0,1)∪(3,∞).$