Find the area of largest rectangle that can be inscribed in an ellipse

The vertices of any rectangle inscribed in an ellipse is given by $$(\pm a \cos(\theta), \pm b \sin(\theta))$$ The area of the rectangle is given by $$A(\theta) = 4ab \cos(\theta) \sin(\theta) = 2ab \sin(2 \theta)$$ Hence, the maximum is when $\sin(2 \theta) = 1$. Hence, the maximum area is when $2\theta = \dfrac{\pi}2$ i.e. $\theta = \dfrac{\pi}4$. The maximum area is $$A = 2ab$$

Suppose that the upper righthand corner of the rectangle is at the point $\langle x,y\rangle$. Then you know that the area of the rectangle is, as you say, $4xy$, and you know that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;.\tag{1}$$

Thinking of the area as a function of $x$, we have $$\frac{dA}{dx}=4x\frac{dy}{dx}+4y\;.$$ Differentiating $(1)$ with respect to $x$, we have

$$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\;,$$ so $$\frac{dy}{dx}=-\frac{b^2x}{a^2y}\;,$$ and $$\frac{dA}{dx}=4y-\frac{4b^2x^2}{a^2y}\;.$$

Setting this to $0$ and simplifying, we have $y^2=\dfrac{b^2x^2}{a^2}$. From $(1)$ we know that $$y^2=b^2-\frac{b^2x^2}{a^2}\;.$$ Thus, $y^2=b^2-y^2$, $2y^2=b^2$, and $\dfrac{y^2}{b^2}=\dfrac12$. Clearly, then, $\dfrac{x^2}{a^2}=\dfrac12$ as well, and the area is maximized when

$$x=\frac{a}{\sqrt2}=\frac{a\sqrt2}2\quad\text{and}\quad y=\frac{b}{\sqrt2}=\frac{b\sqrt2}2\;.$$

$${1=\frac{{ x }^{ 2 }}{{ a }^{ 2 }} + \frac {{ y }^{ 2 }} {{ b }^{ 2 }}} \ge \frac{2 { xy }}{{ ab }} $$

when and only when $${ x }/{ a } = { y }/{ b },$$ the max is got

i.e. :max of $xy =ab/2$, so $4xy=2ab$.