1. We need to evaluate $I(x)=\int e^{\sin^2 x}(\cos x\sin 2x-\sin x)\,dx,$ with the condition $I(0)=1.$

So first, simplify the integrand.

2. Use $\sin 2x=2\sin x\cos x.$ Then $\cos x\sin 2x=\cos x(2\sin x\cos x)=2\sin x\cos^2 x.$ Hence, $\cos x\sin 2x-\sin x=2\sin x\cos^2 x-\sin x.$ Factor out $\sin x$: $=\sin x(2\cos^2 x-1).$ Now use $2\cos^2 x-1=\cos 2x=1-2\sin^2 x.$ Thus, $\cos x\sin 2x-\sin x=\sin x(1-2\sin^2 x).$ So the integrand becomes $e^{\sin^2 x}\sin x(1-2\sin^2 x).$

3. Substitute $t=\cos x \implies dt=-\sin x\,dx.$ Also, $\sin^2 x=1-\cos^2 x=1-t^2.$ Therefore, \begin{align*} I(x) &= \int e^{1-t^2}(1-2(1-t^2))\sin x,dx \ &= \int e^{1-t^2}(2t^2-1)\sin x,dx. \end{align*} Since $dt=-\sin x\,dx$, $I(x)= -\int e^{1-t^2}(2t^2-1)\,dt.$ Rewrite: $I(x)=\int e^{1-t^2}(1-2t^2)\,dt.$

4. Now observe that $\frac{d}{dt}\bigl(te^{1-t^2}\bigr)=e^{1-t^2}+t\cdot e^{1-t^2}(-2t)=e^{1-t^2}(1-2t^2).$ Hence, $I(x)=te^{1-t^2}+C.$ Substituting back $t=\cos x$, $I(x)=\cos x\,e^{\sin^2 x}+C.$

5. Use the condition $I(0)=1$. Since $\cos 0=1, \qquad \sin^2 0=0,$ we get $I(0)=1\cdot e^0+C=1+C.$ Given $I(0)=1$, $1+C=1 \implies C=0.$ So $I(x)=\cos x\,e^{\sin^2 x}.$

6. Now evaluate at $x=\frac{\pi}{3}$: $\cos\frac{\pi}{3}=\frac12,$ and $\sin^2\frac{\pi}{3}=\left(\frac{\sqrt3}{2}\right)^2=\frac34.$ Therefore, $I\left(\frac{\pi}{3}\right)=\frac12 e^{3/4}.$

7. Comparing with the options:

• A: $-e^{3/4}$
• B: $-\frac12 e^{3/4}$
• C: $e^{3/4}$
• D: $\frac12 e^{3/4}$

So the correct option is $\boxed{D}.$