If \(x+y=\tan ^{-1} y\) and \(\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}\), then \(f(y)\) is equal to

Given, \(x+y=\tan ^{-1} y\)
On differentiating w.r.t. \(x\), we get
\(\begin{array}{ll}
& 1+\frac{d y}{d x}=\frac{1}{1+y^{2}} \cdot \frac{d y}{d x} \\
\Rightarrow \quad & \left(1-\frac{1}{1+y^{2}}\right) \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{y^{2}}{1+y^{2}} \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{d y}{d x}=-\left(\frac{1+y^{2}}{y^{2}}\right)=-1-\frac{1}{y^{2}}
\end{array}\)
Again, differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=0+\frac{2}{y^{3}} \frac{d y}{d x} \\
\Rightarrow \quad \frac{d^{2} y}{d x^{2}} &=\frac{2}{y^{3}} \cdot \frac{d y}{d x} \text { but given, } \\
\frac{d^{2} y}{d x^{2}} &=f(y) \frac{d y}{d x}
\end{aligned}\)
On comparing, we get
\(f(y)=\frac{2}{y^{3}}\)