If ${\mathrm{x}}^2\mathrm{y}=2\mathrm{x}-\mathrm{y}$ then ${\mathrm{y}}^2+\frac{{\mathrm{y}}^4}{2}+\frac{{\mathrm{y}}^6}{3}+\dots \dots \infty =$

$\begin{array}{rcl}{\mathrm{x}}^2\mathrm{y}& =& 2\mathrm{x}-\mathrm{y} \Rightarrow \mathrm{y}(1+{\mathrm{x}}^2)=2\mathrm{x}\end{array}$
$\begin{array}{rcl} & \Rightarrow & \mathrm{y}=\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\end{array}$
$\begin{array}{rcl} & & \\ & & {\mathrm{y}}^2+\frac{{\left({\mathrm{y}}^2\right)}^2}{2}+\frac{{\left({\mathrm{y}}^2\right)}^3}{3}+......\infty \\ & =& -\log {(1-\mathrm{y})}^2 =-\log \left(1-\frac{{\left(2\mathrm{x}\right)}^2}{{(1+{\mathrm{x}}^2)}^2}\right)\\ & =& -\log \left(\frac{{(1-{\mathrm{x}}^2)}^2}{{(1+{\mathrm{x}}^2)}^2}\right)\\ & =& -2\log \left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)\\ & =& 2\log \left(\frac{1+{\mathrm{x}}^2}{1-{\mathrm{x}}^2}\right)\end{array}$