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.
When you solve for $\lambda$ using the systems:
$\begin{cases} 2x = \lambda 4x^3 & (1) \\
2y = \lambda 4y^3 & (2) \\
x^4 + y^4 = 18 & (3)
\end{cases}$
You cancel out $x$ in (1) to get $x^2 = 1/(2\lambda)$, you missed a case that $x=0$.
If $x=0$, then $y^4 = 18$, yields $y = \pm {18}^{\frac{1}{4}}$.
The minimum value of $f(x,y)$ would be $\sqrt{18}$, occurs at $(0,\pm {18}^{\frac{1}{4}})$ or $(\pm {18}^{\frac{1}{4}},0)$.
Best Answer
If you note that $x\neq 0$ and $y\neq 0$ you can divide the second equation by the third and get:
$$\frac{2ye^{2xy}}{2xe^{2xy}} = \frac{2\lambda x}{2\lambda y} \ ,$$ which after simplification says that $x^2=y^2$. Taking the first equation, you can then write $$x^2 + y^2 = 2x^2 = 16$$ This gives $x=\pm 2\sqrt{2}$ and therefore four possible solutions $(x,y)=(\pm 2\sqrt{2},\pm 2\sqrt{2})$, where all sign combinations are possible.