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

Separate variables

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

Integrate both side

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

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

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

Since curve pass through $\left(0, 1\right)$.

$\Rightarrow 1-\frac{1}{1}=\ln \left(e^0+1\right)+C \Rightarrow C=-\ln \left(2\right)$

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

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

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