My math course comes back to multivariate differentiations and dives right into it without a proper refresher. could someone help me on the road on how to crack this problem?
Close to the point (2,0), the curve $x^2+xy+y^2=4$ defines y as an implicit function of x.
Which points on the curve have a horizontal tangent?
We have that
$$x^2+xy+y^2=4 \implies 2xdx+ydx+xdy+2ydy=0\implies(x+2y)dy=-(2x+y)dx \implies \frac{dy}{dx}=-\frac{2x+y}{x+2y}=0 \implies y=-2x$$
thus
$$x^2+x(-2x)+(-2x)^2=4\implies 3x^2=4 \implies x=\pm\frac{2\sqrt 3}{3} \quad y=\mp\frac{4\sqrt 3}{3}$$
As an alternative find the explicit form
$$y(x)=\frac{-x\pm\sqrt{16-3x^2}}{2}$$
and evaluate $y'(x)=0$.