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.
1) You really should have simplified your conditions first by dividing out 2 and 4 before differentiating it. This does not change the result, but it is needless complication.
2) $x$, $y$ and $z$ are positive numbers, so we have to take into account the behaviour on the border when one or more of them are zero.
The situation is symmetric, so say $z=0$, but then you have $x+y=50$ and $xy= 750$ which gives (by the solution formula of the quadratic equation) that $x$ and $y$ are not real numbers.
This means that subject to the two conditions there is actually no border and every extrema can be found by using Lagrange multipliers.
3)
You write:
$x(y−z)/2=λ(y−z)$, so that $λ=x/2$
But what you get, for each of the three similar equations, is:
Either (a) $y=z$ or (b) $\lambda=x/2$.
Either (a) $z=x$ or (b) $\lambda=y/2$.
Either (a) $x=y$ or (b) $\lambda=z/2$.
Now, you would have to choose (a) or (b) for each of the three equations which gives 8 cases (aaa,aab,aba,abb,baa,bab,bba,bbb), but it is simpler than that.
If you choose at least once condition (a), then you have two equal variables. If you choose at least twice condition (b), then you have two variables equal to $2\lambda$, so you also have two equal variables.
So, whatever you do, two variables will be equal.
But the problem statement is symmetric in $x$, $y$ and $z$ (which means that if you interchange the three variables, the problem statement does not change, for example $yzx=xyz$, so the volume you are looking for will not depend on the order of the variables).
Therefore, since you know that two variables are equal and you know that it doesn't really matter for the end result which ones are equal, it suffices to investigate $x=y$:
(At this point, we still might have $x=y=z$, but we don't know that. What we do know, is that the original conditions are true, so we use them first, and it turns out that they already fix all the values.)
The original conditions give:
$2x+z=50$ and $x^2+2xz=750$.
Substituting $z=50-2x$ in the second equation gives a quadratic equation $3x^2-100x+750=0$.
This gives $x=\frac{50\pm 5\sqrt{10}}{3}$.
You find two solutions (up to symmetry):
$x=y=\frac53 (10 + \sqrt{10})$, $z=\frac{10}3(5-\sqrt{10})$ giving the volume
$\frac{2500}{27}(35-\sqrt{10})\approx 2948$
and
$x=y=\frac 53 (10- \sqrt{10})$, $z=\frac{10}3(5+\sqrt{10})$ giving the volume
$\frac{2500}{27}(35+\sqrt{10})\approx 3534$.
Since these are the only two candidates for extrema, the first one is the minimum value and the second one is the maximum value.
Best Answer
This proof uses only AM-GM inequality and not Lagrange multipliers. \begin{align*} 2 = \frac{x^2+2y^2+3z^2}{3} \geq (x^2\cdot 2y^2 \cdot 3z^2)^{1/3} = (6x^2y^2z^2)^{1/3} \end{align*} Hence \begin{align*} x^2y^2z^2 \leq \frac{8}{6} \end{align*} and hence \begin{align*} |xyz| \leq \frac{2}{\sqrt{3}} \end{align*}