For the curve $$y = 3\sin\theta\cos\theta$$, $$x = e^\theta\sin\theta$$, $$0 \leq \theta \leq \pi$$, the tangent is parallel to x-axis when $$\theta$$ is:

We begin with the parametric equations of the curve:

$$x = e^{\theta}\sin\theta,$$

$$y = 3\sin\theta\cos\theta,$$

with $$0 \leq \theta \leq \pi.$$

For any curve given in parametric form $$x(\theta),\;y(\theta),$$ the slope of the tangent is obtained from the relation

$$\frac{dy}{dx}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}.$$

A tangent is parallel to the $$x$$-axis when its slope is zero, that is, when $$\dfrac{dy}{dx}=0.$$ This requires

$$\frac{dy}{d\theta}=0 \quad\text{and}\quad \frac{dx}{d\theta}\neq 0.$$

We first compute $$\dfrac{dy}{d\theta}.$$ We have $$y = 3\sin\theta\cos\theta.$$ Using the product rule $$\dfrac{d}{d\theta}\bigl(\sin\theta\cos\theta\bigr)=\cos\theta\cos\theta + \sin\theta(-\sin\theta),$$ we obtain

$$\frac{dy}{d\theta}=3\left(\cos^2\theta-\sin^2\theta\right).$$

Recognising the double-angle identity $$\cos^2\theta-\sin^2\theta=\cos2\theta,$$ this simplifies to

$$\frac{dy}{d\theta}=3\cos2\theta.$$

Next we calculate $$\dfrac{dx}{d\theta}.$$ Because $$x = e^{\theta}\sin\theta,$$ we again apply the product rule, writing

$$\frac{dx}{d\theta}=e^{\theta}\sin\theta + e^{\theta}\cos\theta = e^{\theta}\left(\sin\theta+\cos\theta\right).$$

Now we form the slope of the tangent:

$$\frac{dy}{dx}=\frac{\,3\cos2\theta\,}{\,e^{\theta}\left(\sin\theta+\cos\theta\right)}.$$

The factor $$3/e^{\theta}$$ can never be zero, so the slope vanishes exactly when

$$\cos2\theta = 0.$$

We solve $$\cos2\theta=0.$$ The general solution is

$$2\theta=\frac{(2n+1)\pi}{2}\quad\Longrightarrow\quad\theta=\frac{(2n+1)\pi}{4},\qquad n\in\mathbb{Z}.$$

Restricting to the interval $$0\le\theta\le\pi,$$ we list the admissible values:

For $$n=0:\;\theta=\frac{\pi}{4};$$ for $$n=1:\;\theta=\frac{3\pi}{4};$$ for $$n=2:\;\theta=\frac{5\pi}{4}>\pi,$$ which lies outside the interval and is discarded.

We must still ensure $$\dfrac{dx}{d\theta}\neq0.$$ Evaluating $$\dfrac{dx}{d\theta}=e^{\theta}\bigl(\sin\theta+\cos\theta\bigr)$$ at the two candidates gives

At $$\theta=\frac{\pi}{4}:$$ $$\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2\neq0,$$ so $$\dfrac{dx}{d\theta}\neq0.$$

At $$\theta=\frac{3\pi}{4}:$$ $$\sin\frac{3\pi}{4}+\cos\frac{3\pi}{4}=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=0,$$ hence $$\dfrac{dx}{d\theta}=0,$$ and the expression $$\dfrac{dy}{dx}$$ would be indeterminate, so the tangent is not defined here in the usual sense.

Therefore the only value for which the tangent is parallel to the $$x$$-axis is

$$\theta=\frac{\pi}{4}.$$

Hence, the correct answer is Option C.