Concerning undetermined coefficients method if $f(x)=\sin(2x)\sin(x)$ for the non homogeneous higher order ODE $$y''-3y'+2y=\sin(2x)\sin(x) $$ if we want to solve using undetermined coefficients , can we assume that the particular solution has the form $$y_p(x)=(A\sin(2x)+B\cos(2x))(C\sin(x)+D\cos(x))$$ ? if no then why we can not?

No, you can not assume that you get such a factorization. However, you can use trigonometric identities to get $$ \sin(x)\sin(2x)=\frac12(\cos(x)-\cos(3x)) $$ and compute the particular solutions for each term separately.

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Your friendly helper WolframAlpha gives the solution as follows, which one can then try to factorize, \begin{align} y(x) &= c_1 e^x + c_2 e^{2 x} - \frac{3}{20} \sin(x) + \frac{9}{260} \sin(3 x) + \frac1{20}\cos(x) + \frac{7}{260} \cos(3 x)\\\ &=c_1 e^x + c_2 e^{2 x} \\\&\qquad- \frac{30}{520}\sin(2x)\cos(x)+\frac{48}{520}\cos(2x)\sin(x)+\frac{20}{520}\cos(2x)\cos(x)+\frac{7}{520}\sin(2x)\sin(x)\\\ &=c_1 e^x + c_2 e^{2 x} \\\&\qquad+ \frac{10}{520}(-3\sin(2x)+2\cos(2x))\cos(x)+\frac{6}{520}(8\cos(2x)+\sin(2x))\sin(x) \end{align} at which point we see that the two factors are no further compatible.