Find the tangent plane to the equation $z=2 y \cos (6 x-3 y)$ at the point $(3,6,12)$. \[ z= \]

Step 1 ：We are given the function \(z=2 y \cos (6 x-3 y)\) and we are asked to find the tangent plane at the point \((3,6,12)\).

Step 2 ：The equation of the tangent plane is given by \(z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively, and \((x_0, y_0, z_0)\) is the point of tangency.

Step 3 ：First, we find the partial derivatives of the function \(z(x,y)\) with respect to \(x\) and \(y\).

Step 4 ：The partial derivative of \(z\) with respect to \(x\) is \(z_x = -12y\sin(6x - 3y)\).

Step 5 ：The partial derivative of \(z\) with respect to \(y\) is \(z_y = 6y\sin(6x - 3y) + 2\cos(6x - 3y)\).

Step 6 ：Next, we evaluate these derivatives at the point \((3,6)\).

Step 7 ：We find that \(z_x(3,6) = 0\) and \(z_y(3,6) = 2\).

Step 8 ：Substituting these values into the equation of the tangent plane, we get \(z - 12 = 0*(x - 3) + 2*(y - 6)\).

Step 9 ：Simplifying this equation, we find that the equation of the tangent plane is \(z = 2y\).

Step 10 ：Final Answer: The equation of the tangent plane to the surface defined by the equation \(z=2 y \cos (6 x-3 y)\) at the point \((3,6,12)\) is \(\boxed{z = 2y}\).