How do I apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$? I want to apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$. I'm not able to do it in the standard way, because one term has $z^{-1}$ term and the other has $z$. What is the approach in problems like this?

I think this is about Z-Transformation.
In order to make the overall transfer function causal, we rewrite the given transfer function as
$$ \dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})}. $$
Next, do:
$$ \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})} = \dfrac{A}{a+bz^{-1}} + \dfrac{B}{y+xz^{-1}} $$
It is easy to see that:
$$
aA + yB = 0 \\
xA + bB = K
$$
Solution of this equation set gives:
$$
A = \dfrac{Ky}{xy - ab} \\
B = \dfrac{Ka}{ab - xy}
$$
Therefore, the partial fraction expression is (in the standard first order Z-Transform format):
$$
\dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{\dfrac{Ky}{xy - ab}}{a+bz^{-1}} + \dfrac{\dfrac{Ka}{ab - xy}}{y+xz^{-1}} = \dfrac{\dfrac{Ky}{a(xy - ab)}}{1+\dfrac{b}{a}z^{-1}} + \dfrac{\dfrac{Ka}{y(ab - xy)}}{1+\dfrac{x}{y}z^{-1}}
$$
And the corresponding time domain expression of the given discrete time transfer function is:
$$
f[n] = \left[\dfrac{Ky}{a(xy - ab)} e^{-\frac{b}{a}n} + \dfrac{Ka}{y(ab - xy)} e^{-\frac{x}{y}n}\right] u[n]
$$
