Find the points on the ellipse $4 x^{2}+2 y^{2}=1$ where $f(x, y)=x y$ has its extreme values.
The points are
(Type an ordered pair. Use a comma to separate answers as needed. Type an exact answer, using radicals as needed.)

Step 1 ：We are given the function \(f(x, y) = xy\) and the constraint \(4x^2 + 2y^2 = 1\). We want to find the points on the ellipse where the function has its extreme values.

Step 2 ：We use the method of Lagrange multipliers. This involves setting up a system of equations based on the gradient of the function and the gradient of the constraint.

Step 3 ：The gradient of the function \(f(x, y) = xy\) is \((y, x)\), and the gradient of the constraint \(4x^2 + 2y^2 = 1\) is \((8x, 4y)\).

Step 4 ：We set these equal to each other to get the system of equations: \(y = 8x\lambda\) and \(x = 4y\lambda\). We also have the constraint equation \(4x^2 + 2y^2 = 1\).

Step 5 ：Solving this system of equations gives us the points \((-\sqrt{2}/4, 1/2)\), \((\sqrt{2}/4, -1/2)\), \((-\sqrt{2}/4, -1/2)\), and \((\sqrt{2}/4, 1/2)\).

Step 6 ：Final Answer: The points on the ellipse \(4 x^{2}+2 y^{2}=1\) where \(f(x, y)=x y\) has its extreme values are \(\boxed{(-\sqrt{2}/4, 1/2)}\), \(\boxed{(\sqrt{2}/4, -1/2)}\), \(\boxed{(-\sqrt{2}/4, -1/2)}\), and \(\boxed{(\sqrt{2}/4, 1/2)}\).