The general solution of the differential equation √1+x^2+y^2+x^2 y^2+xy dy dx=0 (where C is a constant of integration)

$$\sqrt{(1+{x}^{2})(1+{y}^{2})}+xy\frac{dy}{dx}=0$$(1+x2)(1+y2)​+xydxdy​=0

Integrating,

$$\Rightarrow \int (\frac{2y}{2\sqrt{1+{y}^{2}}})dy=-\int (\frac{1+{x}^{2}}{x\sqrt{1+{x}^{2}}})dx$$⇒∫(21+y2​2y​)dy=−∫(x1+x2​1+x2​)dx

$$\Rightarrow \sqrt{1+{y}^{2}}=-\int \frac{x}{\sqrt{1+{x}^{2}}}dx-\int \frac{x}{{x}^{2}\sqrt{1+{x}^{2}}}dx$$⇒1+y2​=−∫1+x2​x​dx−∫x21+x2​x​dx

Put $\sqrt{1+{x}^{2}}=t\Rightarrow {x}^{2}={t}^{2}-1$1+x2​=t⇒x2=t2−1 to solve RHS 2nd integration,

$$\Rightarrow \sqrt{1+{y}^{2}}=-\sqrt{1+{x}^{2}}-\int \frac{t}{({t}^{2}-1)t}dt$$⇒1+y2​=−1+x2​−∫(t2−1)tt​dt

$$\Rightarrow \sqrt{1+{y}^{2}}+\sqrt{1+{x}^{2}}=-\int \frac{1}{{t}^{2}-1}dt$$⇒1+y2​+1+x2​=−∫t2−11​dt

$$\Rightarrow \sqrt{1+{y}^{2}}+\sqrt{1+{x}^{2}}=\frac{1}{2}\mathrm{ln}[\frac{\sqrt{1+{x}^{2}}+1}{\sqrt{1+{x}^{2}}-1}]+C$$⇒1+y2​+1+x2​=21​ln[1+x2​−11+x2​+1​]+C