Given: $\left(x\cos \left(\frac{y}{x}\right)\right)\frac{dy}{dx}=y\cos \left(\frac{y}{x}\right)+x$

$\Rightarrow \cos \left(\frac{y}{x}\right)\frac{dy}{dx}=\frac{y}{x}\cos \left(\frac{y}{x}\right)+1$

Putting, $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow \cos v\left(v+x\frac{dv}{dx}\right)=v\cos v+1$

$\Rightarrow v+x\frac{dv}{dx}=v+\frac{1}{\cos v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1}{\cos v}$

$\Rightarrow \left(\cos v\right)dv=\frac{dx}{x}$

$\Rightarrow \int \left(\cos v\right)dv=\int \frac{dx}{x}$

$\Rightarrow \sin v=\log \left|x\right|+c$

$\Rightarrow \sin \left(\frac{y}{x}\right)=\log \left|x\right|+c$

Now, $f\left(1\right)=\frac{\pi }{3}$

$\Rightarrow \sin \left(\frac{\pi }{3}\right)=\log \left|1\right|+c$

$\Rightarrow \frac{\sqrt{3}}{2}=c$

$\Rightarrow \sin \left(\frac{y}{x}\right)=\log \left|x\right|+\frac{\sqrt{3}}{2}$

So, on comparing we get,

$\Rightarrow \alpha =\sqrt{3}$

$\Rightarrow {\alpha }^2=3$