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

To find the equation of the tangent plane to the surface defined by the function \( f(x, y) = \frac{y}{x} \) at the point \( (1, 6, 6) \), we begin by calculating the partial derivatives \( f_x \) and \( f_y \). The partial derivative with respect to \( x \) is: \[ f_x = -\frac{y}{x^2} \] And the partial derivative with respect to \( y \) is: \[ f_y = \frac{1}{x} \] Next, we evaluate these derivatives at the point \( (1, 6) \): \[ f_x(1, 6) = -\frac{6}{1^2} = -6 \] \[ f_y(1, 6) = \frac{1}{1} = 1 \] Now, we can use the point-slope form of the equation of the tangent plane: \[ z - f(1, 6) = f_x(1, 6)(x - 1) + f_y(1, 6)(y - 6) \] Since \( f(1, 6) = 6 \), substituting in the values gives: \[ z - 6 = -6(x - 1) + 1(y - 6) \] \[ z - 6 = -6x + 6 + y - 6 \] \[ z = -6x + y + 6 \] Therefore, the equation of the tangent plane is: \[ z = -6x + y + 6 \]