Let $$f(x) = x - 1$$ and $$g(x) = e^x$$ for $$x \in \mathbb{R}$$. If $$\frac{dy}{dx} = \left(e^{-2\sqrt{x}} g(f(f(x))) - \frac{y}{\sqrt{x}}\right)$$, $$y(0) = 0$$, then $$y(1)$$ is :-

We are given $$f(x)=x-1$$ and $$g(x)=e^{x}$$.

First simplify the composition that appears in the differential equation.

$$f(f(x))=f(x-1)=(x-1)-1=x-2$$
Therefore $$g\!\bigl(f(f(x))\bigr)=e^{\,x-2}$$.

Substituting in the differential equation,

$$\frac{dy}{dx}=e^{-2\sqrt{x}}\;e^{\,x-2}-\frac{y}{\sqrt{x}} \;=\;e^{\,x-2-2\sqrt{x}}\;-\;\frac{y}{\sqrt{x}}.$$

Write it in standard linear form $$\dfrac{dy}{dx}+P(x)\,y=Q(x).$$

Here $$P(x)=\frac{1}{\sqrt{x}},\;\;Q(x)=e^{\,x-2-2\sqrt{x}}.$$

Integrating factor
The integrating factor (I.F.) is $$\exp\!\Bigl(\int P(x)\,dx\Bigr).$$

$$\int P(x)\,dx=\int x^{-1/2}dx=2\sqrt{x},$$
so $$\text{I.F.}=e^{\,2\sqrt{x}}.$$

Multiply the differential equation by this integrating factor:

$$e^{\,2\sqrt{x}}\frac{dy}{dx}+\frac{1}{\sqrt{x}}e^{\,2\sqrt{x}}y =\;e^{\,2\sqrt{x}}\,e^{\,x-2-2\sqrt{x}}=e^{\,x-2}.$$

The left-hand side is the derivative of $$y\,e^{\,2\sqrt{x}}$$, because

$$\frac{d}{dx}\!\bigl(y\,e^{\,2\sqrt{x}}\bigr) =e^{\,2\sqrt{x}}\frac{dy}{dx}+y\,e^{\,2\sqrt{x}}\frac{d}{dx}(2\sqrt{x}) =e^{\,2\sqrt{x}}\frac{dy}{dx}+y\,e^{\,2\sqrt{x}}\frac{1}{\sqrt{x}}.$$

Hence

$$\frac{d}{dx}\!\bigl(y\,e^{\,2\sqrt{x}}\bigr)=e^{\,x-2}.$$

Integrate both sides from $$0$$ to $$x$$:

$$y\,e^{\,2\sqrt{x}}-y(0)\,e^{\,2\sqrt{0}} =\int_{0}^{x}e^{\,t-2}\,dt.$$

Given $$y(0)=0$$ and $$e^{\,2\sqrt{0}}=1$$, so

$$y\,e^{\,2\sqrt{x}} =e^{-2}\!\int_{0}^{x}e^{\,t}\,dt =e^{-2}\bigl(e^{\,x}-1\bigr).$$

Therefore

$$y(x)=e^{-2\sqrt{x}}\;e^{-2}\bigl(e^{\,x}-1\bigr) =e^{\,x-2-2\sqrt{x}}-e^{-2-2\sqrt{x}}.$$

Value at $$x=1$$
For $$x=1$$, $$\sqrt{1}=1$$, so

$$y(1)=e^{\,1-2-2}-e^{-2-2}=e^{-3}-e^{-4}.$$

Factor out $$e^{-4}$$:

$$y(1)=e^{-4}\bigl(e^{\,1}-1\bigr)=\frac{e-1}{e^{4}}.$$

Thus $$y(1)=\dfrac{e-1}{e^{4}}.$$ This matches Option C.