Single digit $\alpha$

$\begin{aligned} y & = \tan^{-1} \left(\frac {4x}{1+5x^2}\right) + \tan^{-1} \left(\frac {2+3x}{2-3x}\right) \\ & = \tan^{-1} \left(\frac {5x-x}{1+5x^2}\right) + \tan^{-1} \left(\frac {\frac 23 +x}{\frac 23-x}\right) \\ & = \tan^{-1} 5x - \tan^{-1} x + \tan^{-1} \frac 23 + \tan^{-1} x \\ & = \tan^{-1} 5x + \tan^{-1} \frac 23 \end{aligned}$

$\implies \dfrac{dy}{dx} = \dfrac 5{1+25x^2}$

Therefore, $\alpha = \boxed 5$ .

$\begin{aligned} \tan y & = \frac{\frac{4x}{1+5x^2}+\frac{2+3x}{3-2x}}{1-\frac{4x}{1+5x^2}\frac{2+3x}{3-2x}} \\ & = \frac{4x(3-2x)+(2+3x)(1+5x^2)}{(1+5x^2)(3-2x)-4x(2+3x)} \\ & = \frac{15x^3+2x^2+15x+2}{-10x^3+3x^2-10x+3} \\ & = \frac{(15x+2)(x^2+1)}{(3-10x)(x^2+1)} \\ & = \frac{15x+2}{3-10x} \end{aligned}$

$\sec ^2 y \frac{dy}{dx}=\frac{15(3-10x)+10(15x+2)}{(3-10x)^2}$

$(1+\tan ^2 y)\frac{dy}{dx}=\frac{65}{9-60x+100x^2}$

$(1+\frac{225x^2+60x+4}{9-60x+100x^2})\frac{dy}{dx}=\frac{65}{9-60x+100x^2}$

$(9-60x+100x^2)+(225x^2+60x+4))\frac{dy}{dx}=65$

$(325x^2+13)\frac{dy}{dx}=65$

$(25x^2+1)\frac{dy}{dx}=5$

$\frac{dy}{dx}=\frac{5}{1+25x^2}$

Therefore $\alpha = \boxed{5}$