\begin{align} & {{x}^{2}}{{y}^{2}}+2y{{x}^{2}}+1=0\Rightarrow {{y}^{2}}+2y+\frac{1}{{{x}^{2}}}=0 \\ & \Rightarrow {{(y+1)}^{2}}=1-\frac{1}{{{x}^{2}}}\overbrace{\Rightarrow }^{x=\sec \theta }y=-1\pm \sin \theta \\ & S=\frac{2}{{{x}^{2}}}+1+\frac{1}{x}+y(y+2+\frac{1}{x}) \\ & y=-1\pm \sin \theta \Rightarrow S=\frac{1}{2}\left( \cos 2\theta \pm \sin 2\theta \right)+\frac{3}{2} \\ & \Rightarrow y=\frac{\sqrt{2}}{2}\cos (2\theta \pm \frac{\pi }{4})+\frac{3}{2} \\ & \Rightarrow \max (y)=\frac{3+\sqrt{2}}{2},\min (y)=\frac{3-\sqrt{2}}{2} \end{align}

I think It is very beautiful problem in Uzbekistan National Olympiad 2013.

Nikpour's answers are correct. I applied the Lagrange multiplier method and got the same results.

Let $x$ and $y$ are real numbers such that $x^2y^2+2yx^2+1=0.$ Prove that $$ -\frac{1}{2}\leq\frac{1}{x^2}+\frac{1}{x}+y(y+2+\frac{1}{x}) \leq \frac{1}{2}$$$$ 1-\frac{3\sqrt 3}{4}\leq \frac{1}{x^2}+\frac{1}{x}+y(y+1+\frac{1}{x})\leq 1+\frac{3\sqrt 3}{4}$$