This shape, which I call the multiplicoid, is the equivalent of, and very similar to, an ellipse. However, instead of the distance between each point and the two focal points summing to a constant, here the distances multiply to give a constant.
Suppose the two focal points are $(-0.5,0)$ and $(0.5,0)$; and the distance of each point on the multiplicoid to each of these points multiplies to give $1$.
$\sqrt {(x+0.5)^2+y^2} * \sqrt {(x-0.5)^2+y^2}= 1$
Neatly, the $x$ intercepts of this shape will occur at $x= \pm \frac{\sqrt 5}{2}$. The y intercepts occur at $y = \pm\frac{\sqrt 3}{2}$
Could the curve of the multiplicoid be expressed as y = $f(x)$? What is its area?
As I noted in a comment to your question, the Cassinian oval is the name given to the curve that is the locus of a point whose distance from two given foci, when multiplied, gives a constant.
In your particular case, the Cassinian oval you have has the rationalized Cartesian equation $$\left(x^2+y^2+\frac14\right)^2-x^2=1$$
Certainly, you can solve for $y$ here, but the expression is not that convenient I think. As for the area, switching to the polar representation and using the formula for the area of a polar curve, along with taking the fourfold symmetry into account, yields
$$\frac14\int_0^{\pi/2}\left(2\cos 2\theta+\sqrt{62+2\cos4\theta}\right)\mathrm d\theta=2E\left(\frac1{16}\right)\approx 3.0919145$$
where $E(m)$ is the complete elliptic integral of the second kind, with parameter $m$.