Square inscribed in a Rhombus.

$d_{1} = \overline{AC} = \sqrt{2}l$ and $d_{2} = 2\overline{BE} = 2\sqrt{1 - \dfrac{l^2}{2}} = \sqrt{2}\sqrt{2 - l^2}$

$A_{ABCD} = \dfrac{1}{2}d_{1}d_{2} = l\sqrt{2 - l^2} \implies A = A_{ABCD} - A_{square} = l\sqrt{2 - l^2} - l^2 \implies$

$\dfrac{dA}{dl} = \dfrac{2(1 - l^2 - l\sqrt{2 - l^2})}{\sqrt{2 - l^2}} = 0 \implies 2l^4 - 4l^2 + 1 = 0$

$l > 1 \implies l = \sqrt{\dfrac{2 - \sqrt{2}}{2}} \implies A = \sqrt{2} - 1$

Note: You can check that a max does occur at $l = \sqrt{\dfrac{2 - \sqrt{2}}{2}}$ .