$$H(z) = \dfrac z {(z-\frac14)(z+\frac14)(z-\frac12)}$$

Poles at: $z=1/4, -1/4, ½$

Pole zero plot:

Since exterior part of circle (r=1/2) lies within the unit circle. Hence the given system is causal. Also all the pole of the system lies inside the unit circle, thus having bounded output for every bounded input. Hence is a stable system

Impulse response:

$H(z) = \dfrac z {(z-\frac14)(z+\frac14)(z-\frac12)} \\ \dfrac {H(z)}z = \dfrac 1 {(z-\frac14)(z+\frac14)(z-\frac12)}$

By using partial fraction,

$\dfrac {H(z)}z = \dfrac A{(z-\frac 14)}+ \dfrac B{(z+\frac14)}+ \dfrac C{(z-\frac12)} $

Put $z=1/4$ we get, $A=-8$

Put $z=-1/4$ we get, $B=2/3$

And Put $z=1/2$ we get, $C=4/3$

$\dfrac {H(z)}z=\dfrac {-8}{(z-\frac14)} + \dfrac {2/3}{(z+\frac14)} + \dfrac {4/3}{(z-\frac12)} \\ H (z) = \dfrac {-8z}{(z-\frac14)} + \dfrac {2z/3}{(z+\frac14 )} + \dfrac {4z/3}{(z-\frac12)} $

By Inverse z-transform,

$ h [n] = -8(\dfrac14)^n u(n)+\dfrac23 (\dfrac {-1}{4})^n u(n)+\dfrac43 (\dfrac12)^n u(n)$