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.
Taking from where you left off. By equating coordinates of both sides, you have: $2z = 4\lambda z\implies z = 0$ or $\lambda = \dfrac{1}{2}$. If $\lambda = \dfrac{1}{2} \implies x = 8$ which is not possible since $x^2 \le 6$. So what is left is $z = 0$ or $\lambda = 0$. Also $\lambda \neq 1$ for otherwise $2x-8 = 2x$ which is not possible. $\lambda \neq 0$, for if it were $0$, then you have: $2x-8 = 0 \implies x = 4$, and this is not possible since $x^2 \le 6$. Thus $2x-8 = 2\lambda x, 2y-4 = 2\lambda y\implies x = \dfrac{4}{1-\lambda}, y = \dfrac{2}{1-\lambda}\implies \dfrac{16}{(1-\lambda)^2}+\dfrac{4}{(1-\lambda)^2}=6\implies (1-\lambda)^2 = \dfrac{20}{6} = \dfrac{10}{3}\implies \lambda = 1\pm \dfrac{\sqrt{30}}{3}$. Can you take it from here ?. It looks as if one of the lambdas will yield a min and the other a max.
Best Answer
Only for this particular problem you have the curve $g(x,y)=x^{2}+y^{2}-2x-2y-23=0$, which is equation of a circle $(x-1)^2+(y-1)^2 = 25$ whose radius is $5$ and center at $(1,1)$. So you can easily calculate the distance.
The distance between the center and the point is $\sqrt{(8-1)^2 + (6-1)^2} = \sqrt{74}$, so the minimum distance is $\sqrt{74}-5$, maximum distance is $\sqrt{74}+5$