EDIT.
I'm inserting here a purely geometrical solution, the original reasoning can be seen at the end.
I'll repeatedly make use of the following result: if we have a line $r$ and an arc of circle $\gamma$, and the point $Q\in\gamma$ which is the farthest from $r$ is not an endpoint of $\gamma$, then the tangent at $Q$ is parallel to $r$.
The vesica piscis is composed of two arcs, with centers $O$ and $O'$. Let's consider the case when the vertices $ABC$ of an inscribed triangle are all internal points of the arcs:
$A$ on arc $O$ and $BC$ on arc $O'$. If $ABC$ is the triangle of maximum area, then $B$ must be the point on the arc which is the farthest from line $AC$ and by the above result the tangent at $B$ must be parallel to $AC$, i.e. radius $O'B$ must be perpendicular to $AC$ and thus lies on an altitude of triangle $ABC$. By the same argument we also get $O'C\perp AB$, hence $O'C$ also lies on an altitude and $O'$ is the orthocenter of the triangle.
But we also have $OA\perp BC$, implying that altitude $OA$ must pass through $O'$, and that can happen only if $A=O'$. Triangle $ABC$ is then isosceles and right angled at $A$, as shown in figure below. Its area is:
$$
area_1={1\over2}d^2.
$$
Let's now consider the case when one of the vertices (e.g. $A$) lies at an endpoint of the arcs, while $B$ and $C$ are internal, each on a different arc as in figure below (it's easy to show that we don't get maximum area if $B$ and $C$ lie on the same arc). Moreover, let $A'$ be the other endpoint and $M$ the midpoint of $OO'$ (see figure below).
For the area to be maximum we need as before $O'B\perp AC$ and
$OC\perp AB$. If we set $\alpha=\angle O'OC$ we can notice that triangles $OFD$ and $AFM$ have both a right angle and a pair of equal vertical angles, hence $\angle MAF=\angle DOF=\alpha$. Moreover, $\angle O'AC={1\over2}\angle O'OC={1\over2}\alpha$, because they are an inscribed and central angle subtending the same arc in circle $O'$.
We can analogously set $\alpha'=\angle OO'B$ to find: $\angle GAM=\alpha'$ and $\angle OAB={1\over2}\alpha'$. But $\angle OAM=\angle O'AM=30°$, hence we have:
$$
\alpha+{1\over2}\alpha'=30°
\quad\text{and}\quad
\alpha'+{1\over2}\alpha=30°.
$$
This system of equation can be easily solved to get:
$$
\alpha=\alpha'=20°.
$$
Hence the largest triangle in this case is isosceles
and symmetric about line $AA'$.
To compute its sides, note that from $\angle ABA'=120°$ we get
$\angle BA'A=60°-\alpha$ and $AB=AC=2d\sin(60°-\alpha)$.
Its area is then
$$
area_2={1\over2}AB\cdot AC\cdot\sin(\alpha+\alpha')=
{1\over2}\big(2d\sin(60°-\alpha)\big)^2\sin2\alpha=
2d^2\sin^340°\approx 0.53 d^2.
$$
Finally, we can consider the case when two points (e.g. $A$ and $B$) lie on the endpoints of the arcs. In that case point $C$ must be either $O$ or $O'$, for the tangent at $C$ to be parallel to $AB$.
The area of triangle $ABC$ is:
$$
area_1={1\over2}d^2\sin120°={\sqrt3\over4}d^2.
$$
In summary, the largest area occurs in the second case.
ORIGINAL REASONING.
Assume WLOG that points $A$ and $B$ are chosen on the right side of the vesica, while $C$ is on the left side (obviously we cannot have maximum area if all three points lie on the same side). For given positions of $A$ and $B$ we get the maximum area if the tangent at $C$ is parallel to $AB$, i.e. if $OC\perp AB$, where $O$ is the center of the right circle (see figure below).
Hence if we choose $C$ at will and some point $D$ on $OC$, points $A$, $B$ of the largest triangle will be the intersections between the left circle (of center $O'$) and the perpendicular to $OC$ at $D$. Note that point $D$ must be close enough to $O$ so that point $A$ won't lie past the intersection point of the circle.
Let $H$ be the projection of $O'$ on $AB$ and $d$ the radius of the circles.
If we set $\alpha=\angle O'OC$ and $x=OD$, then by triangle similitude we get:
$$
O'H:x=\left({d-{x\over\cos\alpha}}\right):{x\over\cos\alpha}
$$
that is:
$$
O'H=d\cos\alpha-x.
$$
We can then find base $AB$ of the triangle:
$$
AB=2\sqrt{d^2-O'H^2}=2\sqrt{d^2-(d\cos\alpha-x)^2}.
$$
The area of triangle $ABC$ is then:
$$
area={1\over2}AB\cdot CD=(d-x)\sqrt{d^2-(d\cos\alpha-x)^2}.
$$
If $x$ is kept constant we get the largest area when $\cos\alpha$ has the lowest possible value, i.e. when $\alpha$ has the greatest possible value (note that $0\le\alpha\le60°$ and $\cos\alpha$ is then positive and decreasing). But the largest possible value is attained by $\alpha$ when point $A$ lies at the intersection of the circles: in the following we can then study only that case.
Let's suppose then that $A$ lies at an intersection of the circles and name $A'$ the other intersection point (see figure below). Note that
$\angle A'AB=\alpha$ and $\angle ABA'=120°$, so that
$\angle AA'B=60°-\alpha$.
We can then compute:
$$
AB=2d\sin(60°-\alpha).
$$
From the equality $AB=2\sqrt{d^2-(d\cos\alpha-x)^2}$ we then get:
$$
x=d\cos\alpha-d\cos(60°-\alpha)
$$
and we we can finally write an expression for the area as a function of $\alpha$:
$$
area={1\over2}AB\cdot (d-x)=d^2\sin(60°-\alpha)
(1-\cos\alpha+\cos(60°-\alpha)).
$$
Expanding, this can be rewritten as:
$$
{area\over d^2}={\sqrt3\over2}\cos\alpha-{1\over2}\sin\alpha +\cos\alpha\sin\alpha-{\sqrt3\over4}.
$$
Now we only need to show that the maximum of this expression is attained for $\alpha=20°$.
Note that it vanishes for $\alpha=60°$ and $\alpha=-120°$, hence there must be an absolute maximum between those values, corresponding to a stationary point.
Differentiating the above expression with respect to $\alpha$ and equating the result to zero, we get the equation:
$$
2\cos^2\alpha-{1\over2}\cos\alpha-1={\sqrt3\over2}\sin\alpha.
$$
Squaring both sides we obtain
$$
(2\cos^2\alpha-{1\over2}\cos\alpha-1)^2={3\over4}(1-\cos^2\alpha),
$$
that is:
$$
16c^4-8c^3-12c^2+4c+1=0,
$$
where I set $c=\cos\alpha$.
This equation has $c={1\over2}$ as solution, which doesn't correspond to an acceptable value for $\alpha$. Hence we can divide the l.h.s. by $2c-1$ to get an equivalent cubic equation:
$$
8c^3-6c-1=0.
$$
But from the cosine triplication formula:
$\cos3x=4\cos^3x-3\cos x$
we get:
$$
4\cos^320°-3\cos20°=\cos60°={1\over2}
$$
implying that $c=\cos20°$ is a solution of the above equation.
The other two solutions can be shown to be negative,
hence the only acceptable stationary point is $\alpha=20°$,
as it was to be proved.
