How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.

As you stated, we must have $x^2 + 3y^2 = 7 $ and $3 = xy + y^2$.
This is an ellipse intersecting a hyperbola, so there are at most 4 points of intersection.
Substtuting $ x = \frac{3-y^2}{y}$ into the first equation, we get a quartic equation
$$ (3-y^2) + 3y^4 = 7y^2. $$
This has solutions $y = -\frac{3}{2}, -1, 1, \frac{3}{2}$.
Hence, this gives us the 4 points of intersection
$$ ( -\frac{1}{2}, - \frac{3}{2} ), ( -2, -1), (2, 1), ( \frac{1}{2}, \frac{3}{2}).$$
