Given,

${\log }_e\left(\frac{dy}{dx}\right)=3x+4y$

$\Rightarrow \frac{dy}{dx}=e^{3x}·e^{4y}$

$\Rightarrow \int e^{-4y}dy=\int e^{3x}dx$

$\Rightarrow \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C$

Given,

$y\left(0\right)=0$

So,

$-\frac{1}{4}-\frac{1}{3}=C\Rightarrow C=-\frac{7}{12}$

So, the particular solution is

$\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}-\frac{7}{12}$

$\Rightarrow e^{-4y}=\frac{4e^{3x}-7}{-3}$

$\Rightarrow e^{4y}=\frac{3}{7-4e^{3x}}\Rightarrow 4y=\ln \left(\frac{3}{7-4e^{3x}}\right)$

$4y=\ell n\left(\frac{3}{6}\right)$ when $x=-\frac{2}{3}\ell n2$

$\Rightarrow y=\frac{1}{4}\ell n\left(\frac{1}{2}\right)$

$\Rightarrow y=-\frac{1}{4}\ell n2$

So, $\alpha =-\frac{1}{4}$