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Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$

(ii) $\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

Given: -

$Y=\left(\tan ^{-1} x\right)^{2}$

Then

$\frac{d y}{d x}=\frac{d\left(\tan ^{-1} x\right)^{2}}{d x}$

Using formula (ii)\&(i)

$\mathrm{y}_{1}=2 \tan ^{-1} \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)$

$\Rightarrow \mathrm{y}_{1}=2 \tan ^{-1} \mathrm{x} \cdot \frac{1}{1+\mathrm{x}^{2}}$

Again differentiating with respect to $x$ on both the sides, we obtain

$\left(1+x^{2}\right) y_{2}+2 x y_{1}=2\left(\frac{1}{1+x^{2}}\right)$ using formula(i)\&(iii)

$\Rightarrow\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$

Hence proved.