Determine the point(s) on the surface $z=x^{2}+6 x+2 y^{2}$ at which the tangent plane is parallel to the $x y$-plane. (Use symbolic notation and fractions where needed. Give your answer as a comma-separated list of coordinate points of the form $(*, *, *)$.)

Step 1 ：Calculate the partial derivatives of \(f(x, y) = x^{2}+6 x+2 y^{2}\), which are \(f_x = 2x + 6\) and \(f_y = 4y\).

Step 2 ：The tangent plane is parallel to the \(xy\)-plane if and only if its normal vector is parallel to the \(z\)-axis. This happens when the coefficients of \(x\) and \(y\) in the equation of the tangent plane are both zero, i.e., \(f_x(x_0, y_0) = 0\) and \(f_y(x_0, y_0) = 0\).

Step 3 ：Setting these equal to zero gives the equations \(2x_0 + 6 = 0\) and \(4y_0 = 0\).

Step 4 ：Solving these equations gives \(x_0 = -3\) and \(y_0 = 0\).

Step 5 ：Substituting these into the equation for the surface gives \(z_0 = (-3)^2 + 6*(-3) + 2*(0)^2 = 9 - 18 = -9\).

Step 6 ：\(\boxed{(-3, 0, -9)}\) is the point on the surface at which the tangent plane is parallel to the \(xy\)-plane.