1. Simplify the given function inside the differential equation

We have $f(x)=x-1, \qquad g(x)=e^x.$

First compute: $f(f(x))=f(x-1)=(x-1)-1=x-2.$ So, $g(f(f(x)))=g(x-2)=e^{x-2}.$

Hence the differential equation becomes $\frac{dy}{dx}=e^{-2\sqrt{x}}\,e^{x-2}-\frac{y}{\sqrt{x}}.$ That is, $\frac{dy}{dx}+\frac{1}{\sqrt{x}}y=e^{x-2\sqrt{x}-2}.$

2. Recognize it as a linear differential equation

This is of the form $\frac{dy}{dx}+P(x)y=Q(x),$ with $P(x)=\frac{1}{\sqrt{x}}, \qquad Q(x)=e^{x-2\sqrt{x}-2}.$

3. Find the integrating factor

The integrating factor is $IF=e^{\int \frac{1}{\sqrt{x}}dx}=e^{2\sqrt{x}}.$

4. Multiply the equation by the integrating factor

Multiplying throughout by $e^{2\sqrt{x}}$: $e^{2\sqrt{x}}\frac{dy}{dx}+\frac{1}{\sqrt{x}}e^{2\sqrt{x}}y=e^{2\sqrt{x}}\cdot e^{x-2\sqrt{x}-2}.$

The right-hand side simplifies to $e^{x-2}.$

The left-hand side becomes $\frac{d}{dx}\left(y e^{2\sqrt{x}}\right).$ Thus, $\frac{d}{dx}\left(y e^{2\sqrt{x}}\right)=e^{x-2}.$

5. Integrate both sides

Integrating, $y e^{2\sqrt{x}}=\int e^{x-2}dx=e^{x-2}+C.$ So, $y=e^{-2\sqrt{x}}\left(e^{x-2}+C\right).$

6. Use the initial condition $y(0)=0$

At $x=0$, $0=e^0\left(e^{-2}+C\right)=e^{-2}+C.$ Hence, $C=-e^{-2}.$

Therefore, $y=e^{-2\sqrt{x}}\left(e^{x-2}-e^{-2}\right).$ Factor out $e^{-2}$: $y=e^{-2\sqrt{x}}e^{-2}(e^x-1).$ So, $y=e^{-2-2\sqrt{x}}(e^x-1).$

7. Find $y(1)$

Substitute $x=1$: $y(1)=e^{-2-2}(e-1)=\frac{e-1}{e^4}.$

8. Match with the options

$\boxed{y(1)=\frac{e-1}{e^4}}$ which is Option B.