3. Find the extreme values of
\[
f(x, y)=x+2 y
\]
on the ellipse
\[
\frac{y^{2}}{2}+\left(\frac{y}{2}\right)^{2}=1
\]

Step 1 ：First, we rewrite the equation of the ellipse in standard form. The given equation is \(\frac{y^{2}}{2}+\left(\frac{y}{2}\right)^{2}=1\). We can rewrite this as \(2y^2 + y^2 = 4\), which simplifies to \(3y^2 = 4\) or \(y^2 = \frac{4}{3}\).

Step 2 ：Next, we find the center of the ellipse. Since the equation is in the form \(y^2 = \frac{4}{3}\), the center of the ellipse is at the origin, (0,0).

Step 3 ：Then, we find the semi-axis in the y-direction. This is given by the square root of the denominator of the y term in the equation of the ellipse, which is \(\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}\).

Step 4 ：Now, we find the extreme values of the function \(f(x, y)=x+2 y\) on the ellipse. The extreme values occur at the vertices of the ellipse. The y-coordinates of the vertices are the y-coordinate of the center plus and minus the semi-axis in the y-direction. So, the y-coordinates of the vertices are \(0 + \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}\) and \(0 - \frac{2}{\sqrt{3}} = -\frac{2}{\sqrt{3}}\).

Step 5 ：Substitute these y-coordinates into the function to find the extreme values. For \(y = \frac{2}{\sqrt{3}}\), we have \(f(x, y) = x + 2(\frac{2}{\sqrt{3}}) = x + \frac{4}{\sqrt{3}}\). For \(y = -\frac{2}{\sqrt{3}}\), we have \(f(x, y) = x - 2(\frac{2}{\sqrt{3}}) = x - \frac{4}{\sqrt{3}}\).

Step 6 ：Finally, we find the maximum and minimum values of these expressions. The maximum value is \(x + \frac{4}{\sqrt{3}}\) and the minimum value is \(x - \frac{4}{\sqrt{3}}\). Therefore, the extreme values of the function on the ellipse are \(\boxed{x + \frac{4}{\sqrt{3}}}\) and \(\boxed{x - \frac{4}{\sqrt{3}}}\).