1. Given differential equation

   $(1+e^{-x})(1+y^2)\frac{dy}{dx}=y^2$

   We need the solution passing through $(0,1)$.

2. Separate variables

   Rearranging,

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

   Since

   $\frac{1+y^2}{y^2}=1+\frac{1}{y^2},$

   we get

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

3. Integrate both sides

   Left side:

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

   Right side:

   $\int \frac{dx}{1+e^{-x}}$

   Multiply numerator and denominator by $e^x$:

   $\frac{1}{1+e^{-x}}=\frac{e^x}{1+e^x}$

   So,

   $\int \frac{dx}{1+e^{-x}} = \int \frac{e^x}{1+e^x}dx = \ln(1+e^x)+C$

   Hence the general solution is

   $y-\frac{1}{y}=\ln(1+e^x)+C$

4. Use the initial condition $(0,1)$

   Substituting $x=0$, $y=1$:

   $1-1=\ln(1+e^0)+C$ $0=\ln 2 + C$ $C=-\ln 2$

   Therefore,

   $y-\frac{1}{y}=\ln(1+e^x)-\ln 2 = \ln\left(\frac{1+e^x}{2}\right)$

5. Rewrite in the form of options

   Multiply both sides by $y$:

   $y^2-1 = y\ln\left(\frac{1+e^x}{2}\right)$

   So,

   $y^2 = 1 + y\ln\left(\frac{1+e^x}{2}\right)$

6. Match with options

   This is exactly Option D:

   $y^2 = 1 + y\log_e\left(\frac{1+e^x}{2}\right)$

Therefore, the correct answer is D.