The points of inflection of a function are given by the zeros of the second derivative. The first derivative of our function is:
[tex]f^{\prime}(x)=2(-\csc\frac{3x}{2}\cot\frac{3x}{2})\cdot\frac{3}{2}=-3(\csc\frac{3x}{2}\cot\frac{3x}{2})[/tex]Using the product rule, the second derivative of our function is:
[tex]\begin{gathered} f^{\prime}^{\prime}(x)=-3\lbrack(\csc\frac{3x}{2})^{\prime}\cdot(\cot\frac{3x}{2})+(\csc\frac{3x}{2})\cdot(\cot\frac{3x}{2})^{\prime}\rbrack \\ \\ =\frac{9}{2}(\csc\frac{3x}{2}\cot^2\frac{3x}{2}+\csc^3\frac{3x}{2}) \end{gathered}[/tex]Then, the zeros of the second derivative are the solutions for the following equation:
[tex]\frac{9}{2}(\csc\frac{3x}{2}\ctg^2\frac{3x}{2}+\csc^3\frac{3x}{2})=0[/tex]Simplifying this expression, we have:
[tex]\begin{gathered} \frac{9}{2}(\csc\frac{3x}{2}\ctg^2\frac{3x}{2}+\csc^3\frac{3x}{2})=0 \\ \\ \csc\frac{3x}{2}\ctg^2\frac{3x}{2}+\csc^3\frac{3x}{2}=0 \\ \\ \ctg^2(\frac{3x}{2})+\csc^2(\frac{3x}{2})=0 \\ \\ \cos^2(\frac{3x}{2})+1=0 \\ \\ \cos^2(\frac{3x}{2})=-1\implies\nexists x\in\mathbb{R} \end{gathered}[/tex]Our function has no inflection points.
