Solution: To solve Laplace's equation $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ with the given conditions, we can separate variables and assume a solution of the form: \[ u(x, y) = X(x)Y(y). \] Substituting this into the equation and dividing by $XY$ yields: \[ \frac{X''}{X} + \frac{Y''}{Y} = 0 \implies -\frac{X''}{X} = \frac{Y''}{Y} = -\lambda, \] where $\lambda$ is a constant. Solving the equation $\frac{Y''}{Y} = -\lambda$ gives: \[ Y'' + \lambda Y = 0 \] The boundary conditions implies that \[ u(x,0) = u(x, \pi ) = 0 \implies Y(0) = Y(\pi ) = 0. \] The above conditions gives $\lambda = n^2$ and the corresponding nonzero solutions are \[ Y_n(x) = \sin (nx). \] The ODE for $X$ with $\lambda = n^2$ is \[ X^{\prime\prime} - n^2 X = 0 \] Solutions are \[ X(x) = X_n(x) = a_n e^{nx} + b_n e^{-nx}. \]

Now using the boundary condition $u = 0$ on the left side of the rectangle (that is $u(0,y) = 0$) implies that $X_n(0) = 0$, that is, $X_n(0) = a_n(0) + b_n(0) = 0 $ or $b_n = -a_n$. This gives \[ X_n(x) = a_n \left( e^{nx} - e^{-nx} \right) = 2a_n \sinh nx. \] Writing the constant $2a_n = A_n$, we obtain the eigenfunction of the given problem \[ u_n(x,y) = A_n \sin ny \sinh nx. \] Therefore, the solution of the given PDE will be \[ u(x,y) = \sum_{n=1}^{\infty} u_n(x,y) = \sum_{n=1}^{\infty} A_n \sin ny \sinh nx. \]

Now we use the another condition that $u_x(\pi ,y ) = \sin y$. \begin{align*} & u_x(x,y) = \sum_{n=1}^{\infty} n A_n \sin ny \cosh nx \\ \implies & \sin y = u_x(\pi ,y) = \sum_{n=1}^{\infty} n A_n \sin ny \cosh n\pi \\ \implies & A_n = \frac{2}{n\pi \cosh n\pi} \int _0^\pi \sin y \sin ny \mathrm{d}y \\ & \kern 0.45cm = \frac{2}{n\pi \cosh n\pi} \begin{dcases} \frac{\pi}{2} , &\text{ if } n=1 ;\\ 0, &\text{ if } \text{ otherwise}. \end{dcases} \\ \implies & A_1 = \frac{1}{\cosh \pi} \text{ and } A_n = 0 \text{ for all } n \geq 2. \end{align*} Therefore, \[ u(x,y) = A_1 \sin y \sinh x = \frac{\sin y \sinh x}{\cosh \pi}. \] Thus, \begin{align*} u\left( x, \frac{\pi}{2}\right) = \frac{\sin \left( \frac{\pi}{2} \right) \sinh x}{\cosh \pi} = \frac{e^{x} - e^{-x}}{e^{\pi}+e^{-\pi }}. \end{align*}