Find an equation of the tangent plane to the surface at the given point. \[ f(x, y)=\frac{y}{x}, \quad(1,6,6) \]

We start with the function \[ f(x,y)=\frac{y}{x}, \] and the point \((1,6,6)\) on the surface, meaning that \( x_0=1 \), \( y_0=6 \), and \( z_0=6 \) since \( f(1,6)=\frac{6}{1}=6 \). The equation of the tangent plane to the surface \( z=f(x,y) \) at the point \((x_0,y_0,z_0)\) is given by \[ z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0). \] 1. **Compute the partial derivatives:** The partial derivative with respect to \( x \) is \[ f_x(x,y)=-\frac{y}{x^2}. \] The partial derivative with respect to \( y \) is \[ f_y(x,y)=\frac{1}{x}. \] 2. **Evaluate the partial derivatives at \((1,6)\):** For \( f_x(1,6) \): \[ f_x(1,6)=-\frac{6}{1^2}=-6. \] For \( f_y(1,6) \): \[ f_y(1,6)=\frac{1}{1}=1. \] 3. **Write the tangent plane equation:** Substitute all values into the tangent plane formula: \[ z-6=-6(x-1)+1(y-6). \] Expanding and simplifying: \[ z-6=-6x+6+y-6, \] \[ z-6=-6x+y, \] \[ z=-6x+y+6. \] Thus, the equation of the tangent plane is: \[ \boxed{z=-6x+y+6.} \]