Find Minima for the equation

Question

Minimum point for $x^2 +2xy + 4y^2 + 6$ is to be found.

$\frac{\partial}{\partial x}$ and equated to $0$ to find $x=-y$

$\frac{\partial}{\partial y}$ and equated to $0$ to find $x = -4y$

I'm confused as to which is the correct solution

Answer 1

$$x^2 +2xy + 4y^2 + 6=(x+y)^2+3y^2+6.$$

The first two terms are nonnegative and can simultaneously achieve zero. This is consistent with your method of taking partial derivatives. Solving your system also leads to $x=y=0$.