1. Rewrite the differential equation

Given $\left[ \frac{x}{\sqrt{x^2-y^2}}+e^{y/x}\right]x\frac{dy}{dx}=x+\left[ \frac{x}{\sqrt{x^2-y^2}}+e^{y/x}\right]y.$

Let $A=\frac{x}{\sqrt{x^2-y^2}}+e^{y/x}.$ Then the equation becomes $A\,x\frac{dy}{dx}=x+A y.$ So, $A\left(x\frac{dy}{dx}-y\right)=x.$

Hence, $x\frac{dy}{dx}-y=\frac{x}{A}.$

2. Use the substitution $\displaystyle v=\frac{y}{x}$

Let $y=vx \quad \Rightarrow \quad \frac{dy}{dx}=v+x\frac{dv}{dx}.$ Thus, $x\frac{dy}{dx}-y=x\left(v+x\frac{dv}{dx}\right)-vx=x^2\frac{dv}{dx}.$

Now simplify $A$: $\sqrt{x^2-y^2}=\sqrt{x^2(1-v^2)}=x\sqrt{1-v^2}$ (since the given points have positive $x$, we take this directly on the relevant curve segment). Therefore, $\frac{x}{\sqrt{x^2-y^2}}=\frac{1}{\sqrt{1-v^2}}.$ So, $A=\frac{1}{\sqrt{1-v^2}}+e^v.$

Hence the DE becomes $x^2\frac{dv}{dx}=\frac{x}{\frac{1}{\sqrt{1-v^2}}+e^v}.$ So, $x\frac{dv}{dx}=\frac{1}{\frac{1}{\sqrt{1-v^2}}+e^v}.$

Equivalently, $\frac{dx}{x}=\left(\frac{1}{\sqrt{1-v^2}}+e^v\right)dv.$

3. Integrate

Integrating both sides, $\int \frac{dx}{x}=\int \left(\frac{1}{\sqrt{1-v^2}}+e^v\right)dv.$ Therefore, $\ln x=\sin^{-1}v+e^v+C.$ Since $v=\dfrac{y}{x}$, $\ln x=\sin^{-1}\left(\frac{y}{x}\right)+e^{y/x}+C.$

4. Use the point $(1,0)$

At $(x,y)=(1,0)$, $\ln 1=\sin^{-1}(0)+e^0+C$ $0=0+1+C$ So, $C=-1.$

Thus the solution is $\ln x=\sin^{-1}\left(\frac{y}{x}\right)+e^{y/x}-1.$

5. Use the point $(2\alpha,\alpha)$

At $(x,y)=(2\alpha,\alpha)$, $\frac{y}{x}=\frac{\alpha}{2\alpha}=\frac12.$ So, $\ln(2\alpha)=\sin^{-1}\left(\frac12\right)+e^{1/2}-1.$ Now, $\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}, \qquad e^{1/2}=\sqrt e.$ Hence, $\ln(2\alpha)=\frac{\pi}{6}+\sqrt e-1.$ Exponentiating, $2\alpha=\exp\left(\frac{\pi}{6}+\sqrt e-1\right).$ Therefore, $\alpha=\frac12\exp\left(\frac{\pi}{6}+\sqrt e-1\right).$

6. Match with the options

This is exactly Option A.