f(x,y)=(4x-x²)cos(y).
∂f/∂x=(4-2x)cos(y) and ∂f/∂y=-(4x-x²)sin(y).
So there is a maximum when 4-2x=0, x=2. The next derivative is negative so this confirms a maximum.
There is a maximum when -sin(y)=0 so y=0 in the given interval.
Therefore there is a maximum when x=2 and y=0, f(x,y)=4.
To find the minimum we need to look at the intervals. When x=1, 4x-x²=3 and when x=3, 4x-x²=3. We know there is a maximum at x=2.
When y=-π/4, cos(y)=√2/2=cos(-y). This is less than cos(0), so is a minimum for y.
We need a minimum for f, so we need x to be 1 or 3 and y to be ±π/4.
There are 4 absolute minima where f=3√2/2 for (x,y)=(1,-π/4), (1,π/4), (3,-π/4), (3,π/4).