$\text{ .If }x+y={\tan }^{-1}y\text{ and }{y}^{11}=f\left(y\right)y^1\text{ then }\mathrm{f}\left(\mathrm{y}\right)=$

$\begin{array}{l}x+y={\tan }^{-1}y\\ 1+y^1=\frac{1}{1+y^2}{y}^1\\ y^{\prime \prime }=\frac{-2y}{{\left(1+y^2\right)}^2}{y}^{\mathrm{\prime }}+\frac{1}{1+y^2}{y}^{\prime \prime }\\ y^{\prime \prime }\left(1-\frac{1}{1+y^2}\right)=\frac{-2y}{{\left(1+y^2\right)}^2}{y}^{\mathrm{\prime }}\\ y^{\prime \prime }\left(\frac{y^2}{1+y^2}\right)=\frac{-2y}{{\left(1+y^2\right)}^2}{y}^{\mathrm{\prime }}\\ y^{\prime \prime }=\frac{-2}{y\left(1+y^2\right)}{y}^{\mathrm{\prime }}\\ \mathrm{f}\left(\mathrm{y}\right)=\frac{-2}{y\left(1+y^2\right)} ( \sin ce y^{\text{'}\text{'}}=f(y\left)y^{\text{'}}\right)\end{array}$