Let the ellipse be in standard position, with (fraction-free) equation
$$b^2 x^2 + a^2 y^2 = a^2 b^2$$
Let our equilateral triangle have circumcenter $(p,q)$ and circumradius $r$. Note that maximizing the area of the triangle is equivalent to maximizing $r$.
For some angle $\theta$ ---actually, for three choices of $\theta$--- the vertices of the triangle have coordinates
$$
(p,q) + r \; \mathrm{cis}\theta \qquad (p,q)+r\;\mathrm{cis}\left(\theta+120^{\circ}\right) \qquad (p,q) + r\;\mathrm{cis}\left(\theta-120^{\circ}\right)
$$
where I abuse the notation "$\mathrm{cis}\theta$" to indicate the vector $(\cos\theta,\sin\theta)$.
Substituting these coordinates into the ellipse equation gives a system of three equations in four parameters $p$, $q$, $r$, $\theta$. I used Mathematica's Resultant[] function to help me eliminate $r$ and $\theta$, arriving at a huge polynomial equation in $p$ and $q$. One factor of the polynomial gives rise to this equation:
$$p^2 b^2\left( a^2+3b^2 \right)^2 + q^2 a^2\left(3a^2+b^2\right)^2= a^2b^2\left(a^2-b^2\right)^2$$
This says that the family of circumcenters $(p,q)$ lie on their own ellipse! We can therefore write
$$p = \frac{a\left(a^2-b^2\right)}{a^2+3b^2}\cos\phi \qquad q = \frac{b\left(a^2-b^2\right)}{3a^2+b^2}\sin\phi$$
for some $\phi$. Back-substituting into the system of equations gives this formula for $r$:
$$r = \frac{4 a b \sqrt{a^2\left(a^2+3b^2\right)^2-\left(a-b\right)^3\left(a+b\right)^3 \cos^2\phi}}{\left(a^2 + 3 b^2\right)\left(3a^2+b^2\right)}$$
The maximum value, $R$, is attained when $\cos\phi = 0$, so
$$R := \frac{4a^2b}{3a^2+b^2}$$
The area of the maximal triangle is
$$\frac{3\sqrt{3}}{4}R^2 = \frac{12a^4 b^2\sqrt{3}}{\left(3a^2+b^2\right)^2}$$
Perhaps-unsurprisingly, the corresponding triangles are centered horizontally within the ellipse, with a vertex at either the top or bottom of the minor axis.
Now, I should point out that my big $pq$ polynomial has other factors, namely, $p$ itself, $q$ itself, and a giant I'll call $f$.
One can verify that the cases $p=0$ and $q=0$ lead to the same results as above. (Specifically, they correspond to the respective cases $\cos\phi=0$ and $\sin\phi=0$.) Intuitively, if a circle's center lies on an axis of the ellipse, then the points of intersection with the ellipse have reflective symmetry over that axis. If there are four distinct points (or two, or none), then we cannot choose three to be the vertices of our equilateral triangle; consequently, there must be only three points of intersection, with one of them on the axis, serving as the point of tangency for the circle and ellipse.
As for the case $f=0$ ... I'll just irresponsibly call it extraneous. (The method of resultants tends to spawn such things.)
If the line has to pass through $(0,b)$, $(3,2)$ and $(a,0)$ it has to satisfy the following equation:
$$
y-0=\dfrac{2-0}{3-a}(x-a)
$$
Observe that when $x=0$, $y=-\dfrac{2}{3-a}a$, that is $b=-\dfrac{2}{3-a}a$
Now if both $a,b$ are non-negative, then certainly the line will have negative slope (why?). Now the area $A(a,b)$ of the triangle formed by the $x$-axis, the $y$-axis and the line, is given by $A(a,b)=\dfrac{1}{2}ab$. By replacing the value of $b$ in this equation we obtain a formula for the area of the triangle as a function on $a$, $A(a)=-\dfrac{1}{2}a\dfrac{2}{3-a}a=\dfrac{a^2}{a-3}$. Now if you want to know for what value of $a$ this function has a maximum or a minimum, differential calculus is of great help.
$$
A'(a)=\dfrac{2a(a-3)-a^2}{(a-3)^2}=-\dfrac{a^2-6a}{(a-3)^2}
$$
To find the critical points of this function we have to find the zeroes of $A'(a)$, that is, we need to find the solutions of the following equation:
$$
-\dfrac{a^2-6a}{(a-3)^2}=0
$$
In this case the zeroes will come from the zeroes of the numerator that is $a^2-6a=0$, but those are $a=0,\quad a=6$, We have to reject $a=0$ since it does not produces a triangle. The answer is $a=6$. So the value of the area will be $A(6)=\dfrac{1}{2}\dfrac{6^2}{(6-3)^2}=2$. In order to find out whether this is a maximum or a minimum you could use the second derivative criteria:
$$
A''(a)=-\dfrac{4(a-3)^3(a^2-6a)-(a-3)^4(2a-6)}{(a-3)^4}=-\dfrac{2(a-3)^3(2a^2-12a-(a-3)(a-3)}{(a-3)^4}=-\dfrac{2a^2-12a-a^2+6a-9}{(a-3)^3}=-\dfrac{a^2-6a-9}{(a-3)^3}
$$
So $A''(6)=-\dfrac{6^2-6(6)-9}{(6-3)^3}=\dfrac{9}{27}=\dfrac{1}{3}>0$, the area is a minimum.
Best Answer
You don't need the secant equation. The area of the triangle is $\frac 12$ base height. The base is $1-x$ as you say. The height is $y$, but you need to express that in terms of $x$. Then you will have area=$f(x)$ and you can differentiate, set the derivative to zero,... It should be clear that the $x$ that maximizes the triangle area will be negative, as when it is positive decreasing it increases both the base and the height.