How do you make a bean?

We can use some known points to rule out options. We know from the graph that there must be three unique real solutions for $y$ when $x=0$ , which are $y=0$ and $y\approx\pm1.8$ ; this immediately rules out three options by inspection, leaving $0.3(x^2+y^2)^2=2x+y^2$ and $0.3(x^2+2y^2)^2=x+4y^2$ as possibilities. We also know that there must be two unique solutions for $x$ when $y=0$ , namely $x=0$ and $x\approx1.85$ . Setting $y=0$ gives us $0.3x^4=2x$ and $0.3x^4=x$ respectively; only the first has $x\approx1.85$ as a solution, so it must be the answer.