1. Interpret the differential equation

   The equation is $\log\left(\frac{dy}{dx}\right)=3x+4y,$ where the logarithm is natural logarithm.

   Exponentiating both sides, $\frac{dy}{dx}=e^{3x+4y}=e^{3x}e^{4y}.$

2. Separate variables

   Rewrite as $e^{-4y}\,dy=e^{3x}\,dx.$

   Integrating both sides, $\int e^{-4y}\,dy=\int e^{3x}\,dx.$

   This gives $-\frac{1}{4}e^{-4y}=\frac{1}{3}e^{3x}+C.$

3. Use the initial condition

   Given $y(0)=0$, $-\frac14 e^0=\frac13 e^0+C$ $-\frac14=\frac13+C$ $C=-\frac14-\frac13=-\frac{7}{12}.$

   So, $-\frac14 e^{-4y}=\frac13 e^{3x}-\frac{7}{12}.$

   Multiply by $-4$: $e^{-4y}=-\frac43 e^{3x}+\frac73.$

   Hence, $e^{-4y}=\frac{7-4e^{3x}}{3}.$

4. Substitute the given value of $x$

   We need $y\left(-\frac{2}{3}\log_e 2\right)$.

   First compute $3x=3\left(-\frac23 \log 2\right)=-2\log 2.$ Therefore, $e^{3x}=e^{-2\log 2}=2^{-2}=\frac14.$

   Substitute into the solution: $e^{-4y}=\frac{7-4\cdot \frac14}{3}=\frac{7-1}{3}=2.$

   So, $-4y=\log 2$ $y=-\frac14 \log 2.$

5. Compare with the given form

   Since $y\left(-\frac23\log_e 2\right)=\alpha \log_e 2,$ we get $\alpha=-\frac14.$

6. Check options

   • A: $-\frac14$ ✅
   • B: $\frac14$ ❌
   • C: $2$ ❌
   • D: $-\frac12$ ❌

Therefore, the correct option is A.