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.
Here is an illustration of the problem at hand.
You want to find the the volume of the tetrahedron formed by the $x=y=z=0$ planes, and the plane tangent to the ellipsoid at some point on its surface, $(x,y,z)$. This volume, of course, will vary with different points on the surface of the ellipse, so the constraint function must be the following equality
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Now the vector normal to the surface of the ellipsoid at some point $(x,y,z)$ is given by the gradient of its equation.
$$\nabla g (x,y,z) = \left<\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}\right>$$
The equation for the plane is tangent to the ellipsoid at this point is easily worked out to be
$$\frac{x}{a^2}x'+\frac{y}{b^2}b'+\frac{z}{c^2}z'=1$$
Now, finally, for the volume that this plane bounds within the first quadrant, we make use of the fact that this volume is simply $1/6$ the volume of the rectangular prism whose side lengths are given by the points of where the plane intersects the three axes.
$$V=\frac{(abc)^2}{6xyz}$$
And that is your function $f$ which you are trying to minimize.
$$f=V$$
$$g\rightarrow \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Best Answer
First, write down the equations you get when differenting $L$. For example, the derivative with respect to $x$ is $$\frac{dL}{dx} = yz + 2\lambda(y+z).$$
Then, you set the derivatives to $0$. This gives you a system of $4$ equations with $4$ variables to solve. Write down that system as an edit to your question so we can see if you made a mistake up to this point.
Edit: As far as I can see, you made no error in calculating the derivatives. Now, I advise you to look at the equations
$$-yz=2\lambda(y+z)\\-xz=2\lambda(x+z)\\-xy=2\lambda(x+y)$$
Try to multiply the first equation by $x$, the second by $y$ and the third by $z$. Play around with what you get, see where that leads you.