If the pair of lines $2x^2+3xy+y^2=0$ makes angles ${\theta }_{1}and{\theta }_2$ with the positive direction of the    X- axis, then

$\begin{array}{l}GivenEquation2x^2+3xy+y^2=0\\ Comparingwitha{x}^2+2hxy+b{y}^2=0\\ a=2,2h=3\Rightarrow h=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,b=1\\ \therefore m_1+m_2=\frac{-2h}{b}=\frac{-3}{1}=-3\\ m_1{m}_2=\frac{a}{b}\Rightarrow \frac{2}{1}\Rightarrow 2\end{array}$

$\begin{array}{l}\mathrm{Tan}({\theta }_1-{\theta }_2)=\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}\\ =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{\sqrt{{(m_1+m_2)}^2-4m_1{m}_2}}{1+m_1{m}_2}\Rightarrow \frac{\sqrt{9-8}}{1+2}\Rightarrow \frac{1}{3}\end{array}$