Absolute extrema on closed boundary I solved a question and want to check whether I did any mistake or not. 
Question. Given $f(x,y)=2y^3 +2x^2y-x^2 -y$, find the absolute min and max over the region $\{x^2+y^2 \le 1\}$. 
First, I check the interior points so $f_x = 4xy-2x = 0 \implies x = 0$ or $y = \frac{1}{2}$. $f_y = 6y^2 + 2x^2 -1 = 0$. Then, to solve these simultaneously, I set  $x = 0$ in $f_y$ so I get 
$$6y^2 -1 = 0 \implies y= \pm \frac{1}{\sqrt{6}}.$$
However, setting $y= \frac{1}{2}$ in $f_y$, I get $\frac{3}{2} + 2x^2 -1 =0 \implies 2x^2 = -\frac{1}{2}$ so that we have no solution. (Is everything okay here ?)
Now, I check the boundary points: $x^2 + y^2 = 1\implies x^2 = 1-y^2$ and plugging this into $f$, we get the following:
$$f(x,y)=2y^3 + 2(1-y^2)y -(1-y^2) -y = y^2 + y -1.$$
To find the extreme points for $f$, $-1\le y \le 1$ and $f' = 2y+1 =0$ must be satisfied so in the end, we get $4$ points:$(0,-1),(\pm \frac{3}{2},-\frac{1}{2}),(0,1)$.
To sum up, there are $6$ critical points in total so $f$ attains its global max $1$ at $(0,1)$ and global min $-\frac{5}{4}$ at the points $(\pm \frac{3}{2},-\frac{1}{2})$.

On the interior, consider the function
$$
\phi(\varepsilon) = f( \varepsilon x' + (1-\varepsilon)x^*)
$$
where $x^*$ is a local maximizer of $f$ on some convex set $K$.  Then a necessary condition is
$$
\phi'(0) = \nabla f(x^*)'(x'-x^*) \le 0, \quad \forall x'\in K
$$
On the interior of the set, like you said, the gradient must vanish since the above condition must hold for all $x'\in K$, so those candidate points are valid.
On the boundary of the feasible set, you basically used the implicit function theorem to derive the standard Lagrangian tangency conditions. The concern is that $x^2 = 1-y^2$ really describes a correspondence and not a function, so substituting and simplifying might be a mistake.  A Lagrangian approach would be to set
$$
\mathcal{L} = 2y^3 + 2x^2 y - x^2 - y - \lambda (x^2+y^2-1)
$$
with necessary conditions
$$
4xy - 2x - \lambda 2x = 0
$$
$$
6y + 2x^2 - 1 - \lambda 2y = 0
$$
$$
-(x^2 + y^2 -1) =0.
$$
When $x=0$, the other conditions give $y^2 = 1$ and $y(6-2\lambda)=1$, with candidate solutions $(x^*,y^*,\lambda^*) = (0,1,5/2)$ and $(x^*,y^*,\lambda^*) = (0,-1,7/2)$. If $y=0$, the system cannot be solved, since it requires $x^2=1$ and $2x^2=1$.
When $x \neq 0$ and $y\neq 0$, the KKT system becomes
$$
4y - 2 - \lambda 2 = 0
$$
$$
6y + 2x^2 - 1 - \lambda 2y = 0
$$
$$
-(x^2 + y^2 -1) =0.
$$
The first two equations imply
$$
2y -1 = \dfrac{6y+2x^2-1}{2y}
$$
or
$$
4y^2 - 8y -2x^2 +1=  0.
$$
If $x^2=1-y^2$, this implies $6y^2-8y-1 = 0$, which doesn't seem to characterize the same critical points as in your answer.
Hope that helps! Sorry if there are any mistakes!
