Consider the function $f(x,y)=3\sqrt{xy}$: how can I find maxima/minima of $f$ on the domain $D=\{(x,y)\in \mathbb{R}^2\mid x\geq 0, y\geq 0, x^2y+y^2+x\leq 3\}$?
Obviously, minima occur on the $x$ and $y$ axis, while maxima (which clearly exist) lie on the curve $x^2y+y^2+x=3$ with $0<x<3$.
First of all, I can consider the function $g(x,y)=xy$ instead of $f$, as they share (I hope) the same points of bounded minima/maxima in the first quadrant. Using Lagrange's multipliers, I get the system $$\left\{\begin{array}{l} y-\lambda xy-\lambda=0\\ x-\lambda x^2-2\lambda y=0\\ x^2y+y^2+x=3 \end{array}\right.$$ I can see that $x=y=1$, $\lambda=\frac{1}{3}$ is a solution, but I'm not able to prove that it is unique for $0<x<3$.
Another possibility is to obtain $y$ from $x$ in the equation of the curve $x^2y+y^2+x=3$, by completing the square I get (for positive $y$) $y=\frac{\sqrt{x^4-4x+12}-x^2}{2}$. Then I tryed to study the sign of the derivative of $h(x)=x\cdot \frac{\sqrt{x^4-4x+12}-x^2}{2}$, again I see that the derivative is $0$ for $x=1$, but I have no clue how about to prove that this is the unique solution.