With $$\tilde{\mathbf{E}}=2j\hat{y}E_me^{-jk_zz}\sin(k_xx)\quad\text{for}\quad 0<x<a$$ the phasor form of Faraday's law $\nabla\times\tilde{\mathbf{E}}=-j\omega\mu_0\tilde{\mathbf{H}}$ leads to $$\tilde{\mathbf{H}}=\frac{1}{-j\omega\mu_0}\begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\\frac{\partial}{\partial x}&0&\frac{\partial}{\partial z}\\0&E_y&0\end{vmatrix}=\frac{-\hat{x}\frac{\partial}{\partial z}E_y+\hat{z}\frac{\partial}{\partial x}E_y}{-j\omega\mu_0}.$$
Could you please explain how to take curl of phasor $\tilde{\mathbf{E}}$? I know calculus 3, but here we have complex numbers and exponential that makes it difficult and confusing.
For example: why some are 0, some are not inside determinant?
Too long for a comment
The vector field $\tilde{\mathbf{E}}$ has only a component in the $\hat{y}$ direction. Assuming $E_m,k_z,k_x$ are constants your $\tilde{\mathbf{E}}$ is of the form $$\tag{1} \tilde{\mathbf{E}}(x,z)=const.\begin{pmatrix}0\\E_y(x,z)\\0\end{pmatrix} $$ where $E_y(x,z)=e^{-jk_zz}\sin(k_xx)\,.$ This specific form of $E_y(x,z)$ does not matter since your final equation for $\tilde{\mathbf{H}}$ does not make use of it. It also does not matter that there is the imaginary unit $j$ in $E_y(x,z)\,.$
Can you calculate the curl of (1) ?