I am trying to solve the problem $\max_{x,y} f(x,y)=2x^3-3x^2 \textrm{ s.t. } g(x,y)=(3-x)^3-y^2=0$ where both $f(x,y)$ and $g(x,y)$ are defined on $\mathbb{R}^2$. I would like to know if the following solution is logically sound.
From the first- and second-order conditions for $f(x,y)$, a local maximum and local minimum exist for $f(x,y)$ at $x=0$ and $x=1$ respectively. Based on the constraint, since $y^2\geq0$, $(3-x)^3 \geq 0$ and therefore $x\leq 3$. We shall then check the boundary point, stationary points and limit to infinity for $x\leq 3$. Because $\lim_{x \to -\infty} f(x,y) = -\infty$, $f(0,y)=0$, $f(2,y)=-1$ and $f(3,y)=27$, $(x,y)=(3,0)$ must maximize $f(x,y)$.
I also notice that the Lagrange method for this problem does not work because it fails the constraint qualification. In particular, the first-order conditions for $L=2x^3-3x^2-\lambda[(3-x)^3-y^2]$ are $\frac{\partial L}{\partial x}=6x^2+3\lambda(3-x)^2=0$ and $\frac{\partial L}{\partial y}=-6y+2\lambda y=0$, and yield no solution. In that case, is checking the points that fail the constraint qualification ($\nabla g(x^*,y^*) \neq (0,0))$ the only way to solve the problem?
Assuming $f(x,y) = 2x^3-3y^2$, the Lagrange method of multipliers is based on regularity to the point of non null gradient existence in the involved functions. This method relies on the possibility of tangency between the objective function level surfaces and the restriction. Those points when they exist are called stationary points. In the present case we have
$$ \cases{ \nabla f = (6x^2,-6y,0)\\ \nabla g = (-3(3-x)^2,-2 y,0) } $$
making now $\nabla f\times\nabla g = (0,0,108x y-162 y-30 x^2 y)$ we can verify that the only point in which $\nabla f = \lambda \nabla g$ is at $y=0$ but at this point $\nabla g=(0,0,0)$ which invalidates the method. Resuming: the method doesn't fail, because it can't be applied here.
NOTE
If $f(x,y) = 2x^3-3x^2$ then
$$ \nabla f\times\nabla g = (0,0,-12x(x-1)y) $$
and we have linear dependency at $x = 0, x = -1, y = 0$. If $y = 0$ then $\nabla g=(0,0,0)$ and if $x=0$ or $x = 1$ we have $\nabla f=(0,0,0)$