The idea espoused in this answer works here too.
$$(x^2+y^2-1)^3=x^2y^3\implies y=\frac{x^{2/3}\pm\sqrt{x^{4/3}-4(x^2-1)}}2\,dx$$
$$A=2\int_0^{a^{3/2}}\sqrt{x^{4/3}-4x^2+4}\,dx$$
Substitute $t=x^{2/3}$:
$$A=3\int_0^a\sqrt{t(t^2-4t^3+4)}\,dt=3.66197272581827\dots$$
Here $a=1.09066\dots$ is the only real root of $t^2-4t^3+4$.
While this can theoretically be expressed through complete elliptic integrals (since the integral runs from zero to zero of the quartic), the resulting form is so complicated that it's not worth it. I did it anyway, and this is the result as Mathematica code:
Print[3 NIntegrate[Sqrt[t(t^2-4t^3+4)], {t, 0, Root[t^2-4t^3+4,1]}, WorkingPrecision->50]]
r = Sqrt[Root[x^3+3x^2-1732, 1]]
m = 1/2 + AlgebraicNumber[r, {0, -1/96, 0, -863/83136, 0, 1/83136}]
n = 1/2 + AlgebraicNumber[r, {0, -7/288, 0, -3461/249408, 0, 1/249408}]
Sqrt[r/2]/32 * (3*EllipticE[m] - AlgebraicNumber[r, {2, 48, 5, 0, 2}]/3*EllipticK[m] +
AlgebraicNumber[r, {-69280, 2661218, 138560, 4043, 55424, 1925}]/(3*64*433)*EllipticPi[n, m]) // N[#,50]& // Print
Your area uses $\cos^{-1}(y)$’s series which converges on $|y|<1$ with the Pochhammer symbol $(u)_v$:
$$\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=2\int_0^\frac\pi2\cos^{-1}\left(2\sin^2\left(\frac x2\right)\right)dx=2\int_0^\frac\pi2\frac\pi2-\sum_{n=0}^\infty\frac{\left(2\sin^2\left(\frac x2\right)\right)^{2n+1}\left(\frac12\right)_n}{(2n+1)n!}dx$$
Switching operators gives:
$$\frac{\pi^2}2-\sum_{n=0}^\infty\frac{2^{2n+2}\left(\frac12\right)_n}{(2n+1)n!}\int_0^\frac\pi2 \sin^{4n+2}\left(\frac x2\right)dx= \frac{\pi^2}2-\sum_{n=0}^\infty\frac{2^{2n+1}\left(\frac12\right)_n}{(2n+1)n!}\int_0^\frac\pi4 \sin^{4n+2}(x)dx $$
Next, use the $\int \sin^r(x)$ to incomplete beta B$_x(a,b)$ identity.
$$\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=\frac{\pi^2}2-\sum_{n=0}^\infty\frac{\left(\frac12\right)_k4^{n+1}}{(2n+1)n!}\text B_\frac12\left(\frac32+2n,\frac12\right)$$
Finally, use a B$_x(a,b)$ series, switch the series, and simplify with hypergeometric $_4\text F_3$:
$$\boxed{\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=\frac1{\sqrt\pi}\left(4\Gamma^2\left(\frac34\right)\,_4\text F_3\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;\frac14\right)+\frac{\Gamma^2\left(\frac14\right)}{36}\,_4\text F_3\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;\frac14\right)\right)= 3.64744119225206997400916081676416139843462\dots}$$
Best Answer
Setting $x=r \cos(\theta), y = r \sin(\theta)$ (the classical change of variables : polar $\to$ cartesian) in your equation gives
$$r^4(\underbrace{(\cos(\theta)^2+\sin(\theta)^2)^2}_{=1}-\underbrace{2\cos(\theta)^2.\sin(\theta)^2}_{\tfrac12 \sin(2 \theta)})=r^2 \underbrace{2 \cos(\theta).\sin(\theta)}_{\tfrac12 \sin(2 \theta)}.$$
Simplifying by $r^2$ and taking the square root, one gets :
$$r(\theta)=\sqrt{\dfrac{\sin(2\theta)}{1-\tfrac12\sin(2\theta)^2}} \ \ \ \text{if} \ \ \ \sin{2 \theta} \geq 0$$
The last condition means that the curve will exist if $0 \leq \theta \leq \pi/2$ (first quadrant) and/or $\pi \leq \theta \leq 3\pi/2$ (third quadrant). No part of the curve in the second and fourth quadrants.
It remains to integrate to obtain the area enclosed by the curve:
$$A=\frac12\int_0^{2 \pi}r(\theta)^2 d \theta$$
BUT, this integral is equal to $0$ because we turn once in the positive sense, once in the negative one. We have to take a serious look at the curve :
We are obliged, if we want to have the unsigned area of a "petal" to integrate from $0$ to $\pi/2$ (and afterwards double the result as a final answer). Up to you for the calculations (as you desire mainly hints).