Determine the analytic function $f(z)= u+iv $, if $u - v = {{\cos x + \sin x - {e^{ - y}}} \over {2(\cos x - \cosh y)}}$ and $f({\pi}/{2}) = 0$.
By observation we find $u = {{\cos x + \sin x -1} \over {2(\cos x - \cosh y)}}$ and $v ={{-1 + {e^{ - y}}} \over {2(\cos x - \cosh y)}}$ satisfying the given conditions and the function is given as $f(z)= u+iv $.
Is this answer correct or not? Moreover, what is the systematic way to find the solution for this problem.
$\require{cancel}$ As I mentioned in the comments, given that $f(z)$ is analytic, then the Cauchy-Riemann equations hold, in other words, $$\begin{cases} u_x=v_y\\ u_y=-v_x \end{cases}$$ are true.
Hint:
Let $$g(x,y)=\frac{\cos x + \sin x - {e^{ - y}}}{2(\cos x - \cosh y)}$$ for simplicity purposes. Then $$u-v=g(x,y)$$ and by differentiating with respect to $x$ and $y$, you obtain $$\begin{cases} u_x-v_x=\frac{\partial g(x,y)}{\partial x}\\ u_y-v_y=\frac{\partial g(x,y)}{\partial y}. \end{cases}$$ Now, using the C-R differential equations, you have $$\begin{cases} u_x+u_y=\frac{\partial g(x,y)}{\partial x}\\ u_y-u_x=\frac{\partial g(x,y)}{\partial y} \end{cases}$$ and adding the two, leads to $$u_y=\frac12\left(\frac{\partial g(x,y)}{\partial x}+\frac{\partial g(x,y)}{\partial y}\right)\implies u=\frac12\int\left(\frac{\partial g(x,y)}{\partial x}+\frac{\partial g(x,y)}{\partial y}\right)dy+D(x)$$ Similarly, $$v_x=-\frac12\left(\frac{\partial g(x,y)}{\partial x}+\frac{\partial g(x,y)}{\partial y}\right)\implies v=-\frac12\int\left(\frac{\partial g(x,y)}{\partial x}+\frac{\partial g(x,y)}{\partial y}\right)dx+E(y)$$ After some integration, you obtain $$u=-\frac{e^y \sin (x)}{-2 e^y \cos (x)+e^{2 y}+1}+D(x)+C_1\\ v=-\frac{e^{-y} \left(e^{2 y}-1\right)}{4 (\cos (x)-\cosh (y))}+E(y)+C_2$$
I'll add the rest of the answer. Since the Cauchy-Riemann equations hold, we have that $$u_x=v_y\implies \cancel{\frac{1-\cos (x) \cosh (y)}{2 (\cos (x)-\cosh (y))^2}}+D'(x)=\cancel{\frac{1-\cos (x) \cosh (y)}{2 (\cos (x)-\cosh (y))^2}}+E'(y)\\ \implies D'(x)=E'(y)=\text{Constant } F$$ thus $$u=-\frac{e^y \sin (x)}{-2 e^y \cos (x)+e^{2 y}+1}+Fx+C_1\\ v=-\frac{e^{-y} \left(e^{2 y}-1\right)}{4 (\cos (x)-\cosh (y))}+Fy+C_2$$ and using the fact that $f(\frac\pi2+i0)=0$ leads to $C_2=0$ and $C_1=\frac{1}{2} (1-\pi F)$, and since $u-v=g(x,y)$ needs to hold, then $F=0$. Finally your components are $$\boxed{u=-\frac{e^y \sin (x)}{-2 e^y \cos (x)+e^{2 y}+1}+\frac12\\ v=-\frac{e^{-y} \left(e^{2 y}-1\right)}{4 (\cos (x)-\cosh (y))}}$$
Addendum
Here is a check, using a CAS, that the components found satisfy all the conditions of the problem statement: