The given function is $f(x) = {\left[\log (\sec x + \tan x)\right]^3}$ with domain $D = \left( - {\pi \over 2},{\pi \over 2} \right)$ and codomain $R$.

Step 1: Check if the function is odd or even (Options A and D)

A function is odd if $f(-x) = -f(x)$ for all $x$ in its domain. A function is even if $f(-x) = f(x)$ for all $x$ in its domain. The domain $D = \left( - {\pi \over 2},{\pi \over 2} \right)$ is symmetric about $x=0$, so we can check for these properties.

Let's evaluate $f(-x)$: $f(-x) = {\left[\log (\sec (-x) + \tan (-x))\right]^3}$

Using the properties of trigonometric functions, we know that $\sec(-x) = \sec x$ and $\tan(-x) = -\tan x$. Substituting these in, we get: $f(-x) = {\left[\log (\sec x - \tan x)\right]^3}$

To relate this back to $f(x)$, we use the identity $\sec^2 x - \tan^2 x = 1$. This can be factored as $(\sec x - \tan x)(\sec x + \tan x) = 1$. From this, we have: $\sec x - \tan x = \frac{1}{\sec x + \tan x} = (\sec x + \tan x)^{-1}$

Substitute this expression back into $f(-x)$: $f(-x) = {\left[\log \left((\sec x + \tan x)^{-1}\right)\right]^3}$

Using the logarithm property $\log(a^b) = b \log a$: $f(-x) = {\left[-1 \cdot \log (\sec x + \tan x)\right]^3}$ $f(-x) = (-1)^3 {\left[\log (\sec x + \tan x)\right]^3}$ $f(-x) = - {\left[\log (\sec x + \tan x)\right]^3} = -f(x)$

Since $f(-x) = -f(x)$, the function $f(x)$ is an odd function. Therefore, Option A is correct, and Option D is incorrect.

Step 2: Check if the function is one-one (injective) (Option B)

A function is one-one if it is strictly monotonic on its domain. We can check this by analyzing the sign of its derivative, $f'(x)$.

Let's find the derivative of $f(x)$ using the chain rule: $f'(x) = 3 {\left[\log (\sec x + \tan x)\right]^2} \cdot \frac{d}{dx} \left[\log (\sec x + \tan x)\right]$

First, find the derivative of the inner function, $g(x) = \log (\sec x + \tan x)$: $g'(x) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$ $\frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$ So, $g'(x) = \frac{1}{\sec x + \tan x} \cdot \sec x (\sec x + \tan x) = \sec x$

Now, substitute this back into the expression for $f'(x)$: $f'(x) = 3 {\left[\log (\sec x + \tan x)\right]^2} \cdot (\sec x)$

Let's analyze the sign of $f'(x)$ for $x \in \left( - {\pi \over 2},{\pi \over 2} \right)$:

1. The term ${\left[\log (\sec x + \tan x)\right]^2}$ is a square, so it is always non-negative, i.e., $\ge 0$.
2. For $x \in \left( - {\pi \over 2},{\pi \over 2} \right)$, $\cos x > 0$, so $\sec x = 1/\cos x > 0$.

Thus, $f'(x) \ge 0$ for all $x$ in the domain. The derivative $f'(x)$ is zero only if $\log(\sec x + \tan x) = 0$, which means $\sec x + \tan x = 1$. This occurs at $x=0$. Since $f'(x) \ge 0$ and is zero only at a single point, the function $f(x)$ is strictly increasing on its domain. A strictly increasing function is always one-one. Therefore, Option B is correct.

Step 3: Check if the function is onto (surjective) (Option C)

A function is onto if its range is equal to its codomain. The codomain is given as $R$ (all real numbers). We need to find the range of $f(x)$.

Let's find the range by analyzing the composite function step-by-step:

1. Let $h(x) = \sec x + \tan x$. For $x \in \left( - {\pi \over 2},{\pi \over 2} \right)$.

   • As $x \to \left({\pi \over 2}\right)^-$, $\sec x \to +\infty$ and $\tan x \to +\infty$. So, $h(x) \to +\infty$.
   • As $x \to \left(-{\pi \over 2}\right)^+$, $h(x) = \frac{1+\sin x}{\cos x}$. This is a $0/0$ form. Using L'Hopital's rule, $\lim_{x \to (-{\pi/2})^{+}} \frac{\cos x}{-\sin x} = \frac{0}{-(-1)} = 0$. Since $h(x)$ is continuous and increasing, its range is $(0, \infty)$.

2. Let $g(x) = \log(h(x)) = \log(\sec x + \tan x)$. The domain of $\log$ is $(0, \infty)$, which is the range of $h(x)$. As $h(x)$ ranges from $0$ to $\infty$, $\log(h(x))$ ranges from $\log(0^+) \to -\infty$ to $\log(\infty) \to +\infty$. So, the range of $g(x)$ is $(-\infty, +\infty)$, which is $R$.

3. Our function is $f(x) = [g(x)]^3$. The function $u \mapsto u^3$ is a strictly increasing function from $R$ to $R$. As the input $g(x)$ ranges over all real numbers $R$, the output $[g(x)]^3$ also ranges over all real numbers $R$.

So, the range of $f(x)$ is $R$. Since the codomain is also $R$, the function is onto. Therefore, Option C is correct.

Final Conclusion: Options A, B, and C are correct.