Let $z=z(x,y)$ be implicit function with two variables from $2sin(x+2y-3z)=x+2y-3z$. Find $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$.
1. Consider the mapping \( F(x,y,z(x,y)) := 2\sin(x+2y-3z) - (x+2y-3z) \) and suppose that \( F(x,y,z(x,y)) = 0 \) implicitly defines \( z \) as a continuously differentiable function of \( x \) and \( y \). This means \( F \in \mathcal{C}^{1} \) over its domain.

2. By the Implicit Function Theorem, if \( F_z \neq 0 \), then we can express the partial derivatives of \( z \) with respect to \( x \) and \( y \) as:
   \[
   \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \quad \text{and} \quad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}
   \]

3. Compute the partial derivatives of \( F \):
   \[
   F_x = \frac{\partial}{\partial x} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = 2\cos(x+2y-3z) - 1
   \]
   \[
   F_y = \frac{\partial}{\partial y} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = 4\cos(x+2y-3z) - 2
   \]
   \[
   F_z = \frac{\partial}{\partial z} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = -6\cos(x+2y-3z) + 3
   \]

4. Using these partial derivatives, we find:
   \[
   \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2\cos(x+2y-3z) - 1}{-6\cos(x+2y-3z) + 3} = \frac{-2\cos(x+2y-3z) + 1}{6\cos(x+2y-3z) - 3}
   \]
   \[
   \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{4\cos(x+2y-3z) - 2}{-6\cos(x+2y-3z) + 3} = \frac{-4\cos(x+2y-3z) + 2}{6\cos(x+2y-3z) - 3}
   \]

5. Adding these partial derivatives together:
   \[
   \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \frac{-2\cos(x+2y-3z) + 1}{6\cos(x+2y-3z) - 3} + \frac{-4\cos(x+2y-3z) + 2}{6\cos(x+2y-3z) - 3}
   \]
   \[
   = \frac{(-2\cos(x+2y-3z) + 1) + (-4\cos(x+2y-3z) + 2)}{6\cos(x+2y-3z) - 3}
   \]
   \[
   = \frac{-6\cos(x+2y-3z) + 3}{6\cos(x+2y-3z) - 3}
   \]
   \[
   = \frac{3 - 6\cos(x+2y-3z)}{3 - 6\cos(x+2y-3z)} = 1
   \]

Therefore, we have:
\[
\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 1
\]

The final answer is \(\boxed{1}\)