The maximum value of $\frac{{{{\text{e}}^{\sin {\text{x}}}}}}{{{{\text{e}}^{\cos {\text{x}}}}}}$ is $\text{e}^{\sqrt{2}}$.
To see this, let $y = \frac{{{{\text{e}}^{\sin {\text{x}}}}}}{{{{\text{e}}^{\cos {\text{x}}}}}}$. Then, $y^2 = \text{e}^{\sin x + \cos x} = \text{e}^{\sqrt{2}}$, where the last equality follows from the trigonometric identity $\sin x + \cos x = \sqrt{2} \sin \left( \frac{\pi}{4} + x \right)$. Taking the square root of both sides, we get $y = \text{e}^{\sqrt{2} \sin \left( \frac{\pi}{4} + x \right)}$.
Since $\sin \left( \frac{\pi}{4} + x \right)$ is a periodic function with period $2 \pi$, it takes on all values in the interval $[-1, 1]$. Therefore, $y$ takes on all values in the interval $[1, \text{e}^{\sqrt{2}}]$. Since $y$ is continuous, it must have a maximum value. This maximum value occurs when $\sin \left( \frac{\pi}{4} + x \right) = 1$, which is when $x = \frac{\pi}{4}$. Therefore, the maximum value of $y$ is $\text{e}^{\sqrt{2}}$.
The other options are incorrect because they are not the maximum value of $\frac{{{{\text{e}}^{\sin {\text{x}}}}}}{{{{\text{e}}^{\cos {\text{x}}}}}}$. For example, $1$ is not the maximum value because $y = 1$ when $x = 0$, but $y$ takes on values greater than $1$ when $x$ is not equal to $0$.