Given, hyperbola is $16 x^2-9 y^2=1$
$\frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1$
$\Rightarrow \quad \frac{x^2}{\left(\frac{1}{4}\right)^2}-\frac{y^2}{\left(\frac{1}{3}\right)^2}=1$
We know that
Eccentricity of a hyperbola, $e_1^2=1+\frac{b^2}{a^2}$
$\begin{aligned} & \Rightarrow e_1^2=1+\frac{1 \times 16}{9 \times 1} \\ & \Rightarrow e_1^2=\frac{25}{9}\end{aligned}$
$\Rightarrow \quad e_1=\frac{5}{3}$ ... (i)
Now, eccentricity of its conjugate
$e_2^2=1+\frac{a^2}{b^2}$
$\begin{array}{ll}\Rightarrow & e_2^2=1+\frac{1 \times 9}{16 \times 1} \\ \Rightarrow & e_2^2=\frac{25}{16}\end{array}$
$\Rightarrow \quad e_2=\frac{5}{4}$ ...(ii)
From Eqs. (i) and (ii),
$3 e_1=3 \times \frac{5}{3}=\frac{5}{4} \times 4$
$\Rightarrow \quad 3 e_1=4 e_2 \quad\left[\because e_2=\frac{5}{4}\right]$