Note first that since $L(x,y,z,k) = −xyz−k(xy+2z(x+y)−50) = L(y,x,z,k)$, the problem is symmetric in $x,y$.
Let's proceed with brutal arithmetic. We have 4 equations in 4 unknowns:
$$
\begin{split}
0 &= -L_x = yz + k(y + 2z)\\
0 &= -L_y = xz + k(x + 2z)\\
0 &= -L_z = xy + 2k(x + y)\\
50 &= xy+2z(x+y) \quad (\text{from } c=0)
\end{split}
$$
Claim 1. $x \neq 0$ and $y \neq 0$ and $x + y \neq 0$.
Proof.
Note first that if $x=0$, then the second constraint implies $k=0$ or $z=0$. If $x=k=0$, the first constraint implies $yz=0$ and the last constraint implies $2yz=50$, which is impossible.
Similarly, if $x=z=0$, the first constraint implies $ky = 0$. As we proved, $k=0$ is impossible, so must be $x=y=z=0$, which contradicts the last constraint.
To establish the last claim, note that if $x+y =0$, the third constraint yields $xy=0$ and the last one yields $xy=50$, which is a contradiction.
QED
Now, solving the last constraint for $z$ and the $L_z$ constraint for $k$, we get
$$k = \frac{-xy}{2(x+y)} \text{ and } z = \frac{50-xy}{2(x+y)}.$$
Both of these are well-defined since $x\neq0$ and $y \neq 0$. Plug both of these into the first constraint, getting
$$0 = \frac{y(50-xy)}{2(x+y)}
- \frac{xy^2}{2(x+y)}
- \frac{2xy(50-xy)}{2^2(x+y)^2}$$
and now multiply both sides by $2(x+y)^2 \neq 0$ to get
$$0 = y(50-xy)(x+y) - xy^2(x+y) - xy(50-xy).$$
Divide by $y \neq 0$ and bring the last term into the first term and divide by $y$ again:
$$
\begin{split}
0 &= (50-xy)(x+y) - xy(x+y) - x(50-xy)\\
0 &= y(50-xy) - xy(x+y) \\
0 &= 50-xy - x(x+y) = 50 - 2xy - x^2\\
50 &= x^2 + 2xy
\end{split}
$$
and because the problem is symmetric, the symmetric constraint must hold (if you want, you can derive it the same way from the second constraint):
$$
\begin{split}
x^2 + 2xy &= 50 \\
y^2 + 2xy &= 50
\end{split}
$$
Hence subtracting them yields $x^2 = y^2$, so $x = \pm y$, but $x \neq -y$ by Claim 1, so $x = y$. Now $50 = x^2 + 2xy = 3x^2$ implies $x = \pm \sqrt{50/3} = y$, and there are two solutions: $(\sqrt{50},\sqrt{50})$ and $(-\sqrt{50},-\sqrt{50})$.
Since $x,y$ are lengths, we know only one will make sense: $x = y = \sqrt{50/3}$. Now plug this back in to find $z$ and $k$ as desired.
Best Answer
Both sets of conditions are necessary conditions for a point to be optimal, but they're not quite the same mathematically. The KKT conditions are more restrictive and thus shrink the class of points (from those satisfying the Fritz John conditions) that must be tested for optimality. The additional restriction with KKT is that the Lagrange multiplier on the gradient of the objective function cannot be zero. One of the most important resulting differences is that KKT points for linear programs must be optimal, whereas Fritz John points for linear programs don't have to be.
The section on KKT conditions in Bazarra, Sherali, and Shetty's Nonlinear Programming: Theory and Algorithms (second edition) has a nice discussion of the issues. There are several good examples, especially one in which a Fritz John point for a linear program is shown not to be optimal. That can't happen with KKT points for linear programs.