Solution:

$h(n)=\left(\frac{1}{2}\right)^n u(n) ; x(n)=\left(\frac{3}{4}\right)^n u(n)\\ $

$\begin{aligned} H\left(e^{j \omega}\right)= & \frac{1}{1-\frac{1}{2} e^{-j \omega}} \quad ; \quad X\left(e^{j \omega}\right)=\frac{1}{1-\frac{3}{4} e^{-j \omega}} \\\\ & Y\left(e^{j \omega}\right)=H\left(e^{j \omega}\right) X\left(e^{j \omega}\right)\\ \end{aligned}$

$\begin{aligned} & Y\left(e^{j \omega}\right)=\frac{1}{1-\frac{1}{2} e^{-j \omega}} \cdot \frac{1}{1-\frac{3}{4} e^{-j \omega}}=\frac{e^{j \omega}}{e^{j \omega}-\frac{1}{2}} \cdot \frac{e^{j \omega}}{e^{j \omega}-\frac{3}{4}} \\\\ & \frac{Y\left(e^{j \omega}\right)}{e^{j \omega}}=\frac{e^{j \omega}}{\left(e^{j \omega}-\frac{1}{2}\right)\left(e^{j \omega}-\frac{3}{4}\right)}=\frac{A}{e^{j \omega}-\frac{1}{2}}+\frac{B}{e^{j \omega}-\frac{3}{4}} \\\\ & e^{j \omega}=A\left(e^{j \omega}-\frac{3}{4}\right)+B\left(e^{j \omega}-\frac{1}{2}\right) \end{aligned}\\ $

$\operatorname{At} e^{j \omega}=\frac{3}{4}$

$\frac{3}{4}=B\left(\frac{3}{4}-\frac{1}{2}\right), \quad \therefore B=3\\ $

$\begin{aligned} & \text { At } e^{j \omega}=\frac{1}{2}\\ \\ & \frac{1}{2}=A\left(\frac{1}{2}-\frac{3}{4}\right), \quad \therefore A=-2\\ \end{aligned}\\ $

$\frac{Y\left(e^{j \omega}\right)}{e^{j \omega}}=\frac{-2}{e^{j \omega}-\frac{1}{2}}+\frac{3}{e^{j \omega}-\frac{3}{4}}\\ $

$Y\left(e^{j \omega}\right)=\frac{-2 e^{j \omega}}{e^{j \omega}-\frac{1}{2}}+\frac{3 e^{j \omega}}{e^{j \omega}-\frac{3}{4}}\\ $

Applying IDTFT,

$y(n)=-2\left(\frac{1}{2}\right)^n u(n)+3\left(\frac{3}{4}\right)^n u(n)$