1. Given differential equation

   $y(x+1)\,dx - x^2\,dy = 0$

   with initial condition

   $y(1)=e.$

2. Rewrite in differential form

   From

   $y(x+1)\,dx = x^2\,dy,$

   we get

   $\frac{dy}{dx} = \frac{x+1}{x^2}y.$

   So,

   $\frac{1}{y}\frac{dy}{dx} = \frac{x+1}{x^2} = \frac{1}{x} + \frac{1}{x^2}.$

3. Separate variables and integrate

   $\frac{dy}{y} = \left(\frac{1}{x} + \frac{1}{x^2}\right)dx.$

   Integrating,

   $\int \frac{dy}{y} = \int \left(\frac{1}{x} + \frac{1}{x^2}\right)dx,$

   $\ln |y| = \ln |x| - \frac{1}{x} + C.$

   Therefore,

   $y = Cx e^{-1/x}.$

4. Use the initial condition

   Since $y(1)=e$,

   $e = C(1)e^{-1} = \frac{C}{e}.$

   Hence,

   $C = e^2.$

   So the solution is

   $f(x)= e^2 x e^{-1/x}.$

5. Evaluate the limit as $x\to 0^+$

   We need

   $\lim_{x\to 0^+} e^2 x e^{-1/x}.$

   As $x\to 0^+$, we have $\frac{1}{x}\to +\infty$, so

   $e^{-1/x}\to 0$

   extremely fast. Therefore,

   $x e^{-1/x} \to 0.$

   Hence,

   $\lim_{x\to 0^+} f(x)=0.$

6. Check options

   • A: $e^2$ ❌
   • B: $0$ ✅
   • C: $\dfrac{1}{e^2}$ ❌
   • D: $\dfrac{1}{e}$ ❌

Therefore, the correct option is B.