Step 1: Simplify the formula for $x\ne\tfrac{\pi}{2}$.
Write the bottom as $3\cos^2x=3(1-\sin^2x)$. Factor the top using $1-\sin^3x=(1-\sin x)(1+\sin x+\sin^2x)$:
\[ f(x)=\frac{(1-\sin x)(1+\sin x+\sin^2x)}{3(1-\sin x)(1+\sin x)}. \]
Step 2: Cancel $(1-\sin x)$.
\[ f(x)=\frac{1+\sin x+\sin^2x}{3(1+\sin x)}. \]
Step 3: Check the function joins smoothly.
At $x=\tfrac{\pi}{2}$, $\sin x=1$, giving $\dfrac{1+1+1}{3(1+1)}=\dfrac{3}{6}=\dfrac12$, which is the given value $f\left(\tfrac{\pi}{2}\right)$. So $f$ is continuous there.
Step 4: Differentiate the simplified $f$.
Differentiate $\dfrac{1+\sin x+\sin^2x}{3(1+\sin x)}$ using the quotient rule. The derivative is a smooth expression in $\sin x$ and $\cos x$.
Step 5: Substitute $x=\tfrac{\pi}{2}$.
At $x=\tfrac{\pi}{2}$, $\cos x=0$. Every surviving term carries a factor of $\cos x$, so the whole derivative becomes $0$.
Step 6: State the result.
\[ f'\!\left(\tfrac{\pi}{2}\right)=0. \]
\[ \boxed{0} \]