Given that $\text{'}{m}_1\text{'}$ and $\text{'}{m}_1\text{'}$, $\left(m_1>m_2\right)$ are the slopes of the lines represented by $5x^2-8xy+3y^2=0.$

Given pair of lines passes through origin.

Let $\sqrt{3}y-m_{1}x=0, \sqrt{3}y-m_{2}x=0$ are two lines represented by given equation.

$\Rightarrow \left(\sqrt{3}y-m_{1}x\right)\left(\sqrt{3}y-m_{2}x\right) = 5x^2-8xy+3y^2$

$\Rightarrow 3y^2-xy\sqrt{3}\left(m_1+m_2\right)+m_1{m}_2{x}^2=5x^2-8xy+3y^2$

By comparing coefficients of $x^2, y^2, xy$ on both sides

$m_1+m_2 =\frac{8}{\sqrt{3}}----\left(i\right), m_1{m}_2 =5-----\left(ii\right)$

$m_1-m_2=\sqrt{{\left(m_1+m_2\right)}^2-4m_1{m}_2} = \sqrt{\frac{64}{3}-20}=\frac{2}{\sqrt{3}}$

Let $\frac{m_1}{m_2} = k$

By applying Componendo Dividendo 

$\frac{m_1+m_2}{m_1-m_2}=\frac{k+1}{k-1}\Rightarrow \frac{8}{2}=\frac{k+1}{k-1}$

$\Rightarrow 4=\frac{k+1}{k-1}$

$\Rightarrow 4k-4=k+1$

$\Rightarrow 3k =5$

$\Rightarrow k=\frac{5}{3}$

So required is $\frac{m_1}{m_2}=\frac{5}{3}$.