Step 1: Plan.
Here we will not solve for the constant first. Instead we use a definite integral from the starting point to a general point, which builds the condition in directly.
Step 2: Separate.
From $\frac{dy}{dx}=e^{3x}e^{4y}$ we get $e^{-4y}dy=e^{3x}dx$.
Step 3: Definite integral.
Integrate $y$ from $-\frac{1}{4}$ to $y$ and $x$ from $0$ to $x$.
\[ \int_{-1/4}^{y}e^{-4t}dt=\int_{0}^{x}e^{3s}ds \]
Step 4: Evaluate both sides.
Left side: $\left[-\frac{e^{-4t}}{4}\right]_{-1/4}^{y}=-\frac{e^{-4y}}{4}+\frac{e}{4}$.
Right side: $\left[\frac{e^{3s}}{3}\right]_{0}^{x}=\frac{e^{3x}}{3}-\frac{1}{3}$.
Step 5: Clear the fractions.
\[ -\frac{e^{-4y}}{4}+\frac{e}{4}=\frac{e^{3x}}{3}-\frac{1}{3} \]
Multiply by 12: $-3e^{-4y}+3e=4e^{3x}-4$.
So $3e^{-4y}+4e^{3x}=3e+4$.
Step 6: Match.
This is exactly option 2. Options 1, 3 and 4 fail the point $(0,-\frac{1}{4})$, so they are not solutions of this problem.
Final Answer:
Option 2 is correct. \[ \boxed{3e^{-4y}+4e^{3x}=3e+4} \]