The transformed equation of $2x^3-\frac{3x^2}{2}-\frac{x}{8}+\frac{3}{16}=0$  by eliminating fractional coefficients and having unity for the coefficient of the first term is

Given coefficient of first term will be unity

$f\left(x\right)=x^3-\frac{3}{4}{x}^2-\frac{x}{16}+\frac{3}{32}=0$

Let $f\left(\frac{y}{m}\right)=0$

$f\left(\frac{y}{m}\right)={\left(\frac{y}{m}\right)}^3-\frac{3}{4}{\left(\frac{y}{m}\right)}^2-\frac{1}{16}\left(\frac{y}{m}\right)+\frac{3}{32}=0$

$=y^3-\frac{3}{4}{y}^{2}m-\frac{1}{16}y{m}^2+\frac{3}{32}{m}^3=0$

$\Rightarrow y^3-\frac{3}{2^2}{y}^{2}m-\frac{1}{2^4}y{m}^2+\frac{3}{2^5}{m}^3=0$

Exponent of 2 are 2,4,5

$m=2^2$

$\therefore$ Required equation is $x^3-3x^2-x+6=0$