If $y=(\tan^{-1} x)^{2}$,show that $(x^{2}+1)^{2} y_{2}+2 x(x^{2}+1) y_{1}=2$.

Given: $y=(\tan^{-1} x)^{2}$
Differentiating with respect to $x$:
$y_{1} = 2(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x)$
$y_{1} = 2(\tan^{-1} x) \cdot \frac{1}{1+x^{2}}$
$(1+x^{2}) y_{1} = 2 \tan^{-1} x$
Differentiating again with respect to $x$:
$\frac{d}{dx}[(1+x^{2}) y_{1}] = \frac{d}{dx}[2 \tan^{-1} x]$
$(1+x^{2}) y_{2} + y_{1}(2x) = 2 \cdot \frac{1}{1+x^{2}}$
Multiplying both sides by $(1+x^{2})$:
$(1+x^{2})^{2} y_{2} + 2x(1+x^{2}) y_{1} = 2$
Hence,the result is proved.