Answer
The highest point: $(-2,4)$
The lowest point: $(2,-4)$
Work Step by Step
Differentiating $x^{2}+xy+y^{2}$ = $12$ implicitly with respect to $x$ gives
$2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}$ = $0$
$\frac{dy}{dx}$ = $-\frac{2x+y}{x+2y}$
At a highest or lowest point
$\frac{dy}{dx}$ = $0$
$y$ = $-2x$
Substituting $-2x$ for $y$ in the original equation gives
$x^{2}+x(-2x)+(-2x)^{2}$ = $12$
$3x^{2}$ = $12$
$x$ = $±2$
if $x$ = $2$ then $y$ = $-4$
if $x$ = $-2$ then $y$ = $4$
Thus
the highest point is $(-2,4)$
the lowest point is $(2,-4)$
