Constrained optimization of $f(x,y) = e^{-x^{2}-y} $ 
Let $f(x,y) = e^{-x^{2}-y} $ and the constraint set $M$ be $\{(x,y): y^2 = e^{-x^2}\}$. Then
A. $f(x,y)$ is not bounded on $M$
B. $(0, -1)$ is point of local mimimum
C. $(0,1)$ is point of local maximum.

I simply substituted the constaint into the function, i.e. 
$$f(x,y) = e^{-x^{2}-y} = y^2 e^{-y}  $$
FOC: $$2y e^{-y} - y^2 e^{-y} = 0$$
The critical points are $\{0,2\}$. Hence, B,C are false.
A is true because $\lim_{y \rightarrow - \infty} y^2 e^{-y} = \infty$.
The problem is that the answer is supposed to C. Could you please help me to understand where I made a mistake.
 A: Optimizing $f(x,y) = e^{-x^2 - y}$ on the curve $M : y^2 = e^{-x^2}$, we find that 
$$
f(x,y) = e^{-x^2 - y} = y^2 e^{-y}
. 
$$
Then, $0 = f'(y) = 2y e^{-y} - y^2 e^{-y} = ye^{-y} (2 - y)$, so that $y = 0 , 2$. However, we see that $y=0$ does not lie on $M$, and $y = 2$ corresponds to $x = \sqrt{ - \ln 4 }$, which is imaginary (that is, a non-real coefficient). Hence, neither of these possible critical values for $f$ will work, as they do not correspond to points on the constraint curve $M$!!! (not factorial - emphasis) 
So, we must approach this a little differently - what if we know try to write $f$ as a function of $x$? Well, then we consider the function on the two different paths parametrizing $M$: $f_+$ where $y = (+) \sqrt{e^{-x^2}}$ and $f_{-}$ where $y = (-) \sqrt{e^{-x^2}}$.
$$
f_+ (x) = e^{-x^2 - \sqrt{e^{-x^2}}} , \quad f_- (x) = e^{-x^2 + \sqrt{e^{-x^2}}} .
$$
Now, $f_+ ' (x) = 0 \implies (-2x - \frac{1}{2} \frac{-2x e^{-x^2}}{\sqrt{e^{-x^2}} }) f_+ (x) = 0$, that is, as $|f(x)| > 0$, we must have that $0 = x(\sqrt{e^{-x^2}} -2)$, or $x = 0 ,\sqrt{-\ln 4}$ (and we know which of those we can throw out right away!!!). Note that we can check $f_{-} '( x) = 0$ as well, to find that $0 = -x(2 + \sqrt{e^{-x^2}})$ so that $x = 0 $. Now, we just check that $x=0$ corresponds to a point(s) on the constraint $M$, which it does - the points $(0,1)$ and $(0,-1)$. Now, we just check which of these corresponds to local extrema (and which kind). As we know it must be $B$ or $C$, we can check rather easily which is correct.
