$\text{ If }f\left(x\right)=a{\log }_e\left|x\right|+b{x}^2+x\text{ has extremums at }\mathrm{x}=1\text{ and }\mathrm{x}=3\text{ , then }$

$\begin{array}{l}\text{ at }\mathrm{x}=1,3\text{ we have }\left|x\right|=x\\ \therefore f\left(x\right)=a{\log }_{e}x+b{x}^2+x\\ \therefore f^{\mathrm{\prime }}\left(x\right)=\frac{a}{x}+2bx+1\end{array}$

$\begin{array}{l}\text{ According to question, }{f}^{\mathrm{\prime }}\left(1\right)=0,f^{\mathrm{\prime }}\left(3\right)=0\\ \therefore a+2b+1=0,\frac{a}{3}+6b+1=0\end{array}$

$\text{ On solving, we get }a=\frac{-3}{4},b=-\frac{1}{8}$