Start with the equations that you have derived:
\begin{eqnarray*}
ye^{xy}&=&3\lambda x^2,\\
xe^{xy}&=&3\lambda y^2,\\
x^3+y^3&=&16.
\end{eqnarray*}
As a first step, show that none of $\lambda, x$ or $y$ can be zero. (If one of them is zero, then the first two equations show that all three must be zero, contradicting the third equation.) This is useful as we now know that we can divide by these terms at will.
Now comparing the first and second equations gives $x^3=y^3$, and since both are real, we get $x=y$. The third equation then gives $x=y=2$, and the first or second yields the value of $\lambda$. We find $f=e^4$ at this point, and it must be a maximum.
As regards the minimum, recall that the Lagrange multiplier method identifies the possible location of max/min points IF they exist. I don't think your example has a minimum: By taking $x^3 = N^3$ and $y^3=16-N^3$, where $N$ is a large positive number, the constraint $x^3+y^3=16$ is satisfied. But $xy\sim-N^2$ can be made arbitrarily large and negative, so that $e^{xy}$ can be made as close as we like to $0$, but of course would never equal zero. So we can say that the infimum
$$\inf\{e^{xy}:x^3+y^3=16\}=0,$$
but the minimum
$$\min\{e^{xy}:x^3+y^3=16\}$$
does not exist: for any candidate minimum at $(x_0,y_0)$, we can always find $(x_1,y_1)$ with $0<f(x_1,y_1)<f(x_0,y_0)$.
Another approach is to eliminate $y$ and to treat this as a single variable max/min problem for $h(x)=\exp[x(16-x^3)^{1/3}]$. This gives another way of understanding why there is no minimum point of the function.
While Tom is basically correct that this optimization problem "smells like Snell's", it is missing something significant, which presumably got omitted somewhere in stating the function to be minimized.
The intent of the problem from the poser was likely to show how the description of light refraction referred to as "Snell's Law" comes from a determination of the path for a "ray" of light requiring the least amount of time to pass from a point in one optical medium to another with a different refractive index. A diagram of the physical situation might look like this:
There are two media with a simple interface between them (represented here by a horizontal line). The light ray is traced from a point at a distance $ \ a \ $ "above" the interface to a second point a distance $ \ b \ $ "below" that line.
The differing optical properties of the two media changes the speed and direction of the light ray. The convention is to measure the angle that the ray makes to a "vertical" (normal) line to the interface. If we call $ \ c \ $ the speed of light "in vacuum", and the "refractive indices" of the two media $ \ n_1 \ $ and $ \ n_2 \ $ , then the (constant) speeds of the light ray passing through the two media are $ \ v_1 \ = \ \frac{c}{n_1} \ $ and $ \ v_2 \ = \ \frac{c}{n_2} \ $ , respectively.
The "vertical component" of the velocity of the ray in the medium above the interface is then $ \ v_1 \ \cos \alpha \ $ and in the medium below, $ \ v_2 \ \cos \beta \ $ . The total time required for the ray to travel from the "initial" to "final" point is then
$$ T \ = \ \frac{a}{v_1 \ \cos \alpha} \ + \ \frac{b}{v_2 \ \cos \beta} \ \ . $$
The constraint that is usually imposed in minimizing this "total time function" of the two angle variables $ \ \alpha \ $ and $ \ \beta \ $ is to use the fixed "horizontal" separation between the two endpoints,
$$ L \ = \ a \ \tan \alpha \ + \ b \ \tan \beta \ \ . $$
[Note from the physical arrangement that the angles have "first quadrant" values.]
The Lagrange-multiplier calculation then proceeds as Tom suggests, producing the equations
$$ \frac{a}{v_1} \ \sec \alpha \ \tan \alpha \ = \ \lambda \ \cdot \ a \ \sec^2 \alpha \ \ , \ \ \frac{b}{v_2} \ \sec \beta \ \tan \beta \ = \ \lambda \ \cdot \ b \ \sec^2 \beta $$
$$ \Rightarrow \ \ \lambda \ = \ \frac{1}{v_1} \cdot \frac{\tan \alpha}{\sec \alpha} \ = \ \frac{1}{v_2} \cdot \frac{\tan \beta}{\sec \beta} \ \ \Rightarrow \ \ \frac{\sin \alpha}{v_1} \ = \ \frac{\sin \beta}{v_2} \ \ . $$
If we replace the speeds with the expressions above relating them to the refractive indices, we obtain the familiar expression of "Snell's Law",
$$ \frac{\sin \alpha}{c \ / \ n_1} \ = \ \frac{\sin \beta}{c \ / \ n_2} \ \ \Rightarrow \ \ n_1 \ \sin \alpha \ = \ n_2 \ \sin \beta \ \ . $$
The statement of the problem posted by Drey1 , without the subscripts, would have a single speed for the light ray between the two points, which would then just lead to $ \ \alpha \ = \ \beta \ $ (since the angles are in the first quadrant). The path is then a straight line through an effectively uniform medium, as we would expect.
To answer the question concerning the minimum time for the transit of the light ray, we have an "underdetermined" problem, in the sense that we can only find one of the angles in terms of the other. Since the distances $ \ a \ $ and $ \ b \ $ and the speeds $ \ v_1 \ $ and $ \ v_2 \ $ are set by the physical situation, we can choose the "entry angle" $ \ \alpha \ $ as the "independent variable" to write
$$ \left( \frac{v_2}{v_1} \right)^2 \ \sin^2 \alpha \ = \ \sin^2 \beta \ \ \Rightarrow \ \ \left( \frac{v_2}{v_1} \right)^2 \ \sin^2 \alpha \ = \ 1 \ - \ \cos^2 \beta $$
$$ \Rightarrow \ \ \cos \beta \ = \ \sqrt{ \ 1 \ - \ \left( \frac{v_2}{v_1} \right)^2 \ \sin^2 \alpha} \ \ . $$
Inserting this result into the "total time" for the travel of the light ray yields
$$ T_{min} (\alpha) \ = \ \frac{a}{v_1 \ \cos \alpha} \ + \ \frac{b}{v_2 \ \sqrt{ \ 1 \ - \ \left( \frac{v_2}{v_1} \right)^2 \ \sin^2 \alpha}} \ \ . $$
[There is a limitation to the applicability of this result in passing from one medium to a second where $ \ v_1 \ < \ v_2 \ $ : there will then be a "critical angle" at which the behavior of the ray changes from refraction to "total internal reflection".]
With the two speeds for the ray equal (the "uniform medium" condition), this result simplifies to $ \ T \ = \ \frac{a \ + \ b}{ \cos \alpha} \ $ , which is correct for the "straight-line" path.
Best Answer
From your equations -
$x = \frac {a^2}{2\lambda}, y = \frac {b^2}{2\lambda}, z = \frac {c^2}{2\lambda}$...(i)
Substituting in $\, \frac {x^2}{a^2} + \frac{y^2} {b^2}+ \frac {z^2} {c^2} =1$ ...(ii)
$\frac {a^2}{4\lambda^2} + \frac{b^2} {4\lambda^2}+ \frac {c^2} {4\lambda^2} =1$
$2\lambda = \pm \sqrt {a^2+b^2+c^2}$
Given (ii), extreme values of $x + y + z = \pm\frac{a^2+b^2+c^2}{\sqrt {a^2+b^2+c^2}} = \pm \sqrt {a^2+b^2+c^2}$ (using equation (i)).