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The actual problem reads:
Find the area of the largest rectangle that can be inscribed in the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right.
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 $4xy=2ab$.
let L and H be the length and breadth of the required rectangle respectively
$$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$
$$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$
$$H=\frac{b}{a}\sqrt{4a^2-L^2}$$
Area=L*H
$$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$
$$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0$$
$$\frac{b*(4a^2-2L^2)}{a\sqrt{4a^2-L^2}}$$
$$=> 4a^2-2L^2=0$$
$$=2a^2=L^2$$ $$L=a\sqrt{2}$$ $$H=b\sqrt{2}$$ $$Area=L*H=2ab$$
$$\frac{d^2A}{dL^2}=\frac{\sqrt{4a^2-L^2}*(-4L)-\frac{4a^2-2L^2}{2\sqrt{4a^2-L^2}}}{4a^2-L^2}$$
Putting L=$a\sqrt{2}$ $$\frac{d^2A}{dL^2}=\frac{-a\sqrt{2}(4a\sqrt{2})-\frac{0}{2\sqrt{4a^2-2a^2}}}{4a^2-2a^2}$$
$$=\frac{-8a^2}{2a^2}$$
-4<0.
Therefore the area is max
Stretching the plane in a given direction is an operation that preserves ratios of areas. So:
Stretch the plane by a factor of $a/b$ in the $y$-direction, to transform the ellipse to a circle with radius $a$.
Inscribe the largest possible rectangle inside this circle, which turns out to be a square of area $2a^2$. Align this square with the $xy-$axes.
Stretch the plane by a factor of $b/a$, to return the ellipse to its original size and shape. The resulting rectangle has area $2a^2\cdot b/a = 2ab$.
Parametric form of ellipse
$$ x = a \cos t,\; y = b \sin t, \; A = 4 x y = 2 a b \sin (2 t);$$
EDIT1:
Maximum area occurs for rectangle cutting by radial straight lines at $ t= \pm 45^0 $ through origin. The ellipse area is $\dfrac{2}{\pi}$ fraction of the enveloping rectangle area . The ellipse passes through rectangle corners $ \frac{a}{\sqrt2},\frac{b}{\sqrt2}.$
Due to two axis symmetry slanted orientations of rectangles can be ruled out.
Using the same technique shown here, we can orthogonally project the desired rectangle to the inscribed rectangle in the unit circle with maximal area (i.e. $2$, consider the inscribed square with sidelength $\sqrt{2}$).
Let the maximal area of our rectangle be $\mathcal{A}$. Then, by preservation of area ratios,
$$\begin{align*} \frac{\mathcal{A}}{\text{ Area of ellipse}} &=\frac{\text{Area of maximal square}}{\text{ Area of circle}} \\\frac{\mathcal{A}}{ab\pi} &= \frac{2}{\pi}\\ \implies \mathcal{A} &=2ab\end{align*}$$ which is our answer.
