If $\mathrm{f}(x)=\frac{\sin ^2 x}{1+\cot x}+\frac{\cos ^2 x}{1+\tan x}$, then the value of $\mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

Simplify the function:
$\mathrm{f}(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$. Express $\cot x$ and $\tan x$ in terms of sine and cosine:

$\mathrm{f}(x) = \frac{\sin^2 x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2 x}{1 + \frac{\sin x}{\cos x}} = \frac{\sin^2 x}{\frac{\sin x + \cos x}{\sin x}} + \frac{\cos^2 x}{\frac{\sin x + \cos x}{\cos x}} = \frac{\sin^3 x}{\sin x + \cos x} + \frac{\cos^3 x}{\sin x + \cos x}$.

Combine terms:

$\mathrm{f}(x) = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x}$.

Apply the sum of cubes identity:

$\mathrm{f}(x) = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x}$.

Cancel the common factor (valid for $\sin x + \cos x \neq 0$):

$\mathrm{f}(x) = \sin^2 x + \cos^2 x - \sin x \cos x = 1 - \sin x \cos x$.

Use the double angle identity $\sin(2x) = 2\sin x \cos x$:

$\mathrm{f}(x) = 1 - \frac{1}{2} \sin(2x)$.

Differentiate:
$\mathrm{f}'(x) = \frac{d}{dx} \left(1 - \frac{1}{2} \sin(2x)\right) = -\frac{1}{2} \cdot \cos(2x) \cdot 2 = -\cos(2x)$.

Evaluate at $x = \frac{\pi}{6}$:

$\mathrm{f}'\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$.

The correct option is $\boxed{\text{C}}$.