We have,
$$
\begin{aligned}
& \left(1+y^2\right) d x=\left(\tan ^2 y-x\right) d y \\
\Rightarrow \quad & \left(1+y^2\right) \frac{d x}{d y}=\tan ^2 y-x \\
\Rightarrow & \frac{d x}{d y}+\frac{1}{1+y^2} x=\frac{\tan ^2 y}{1+y^2}
\end{aligned}
$$
which is a linear differential equation
$$
\therefore \quad \text { IF }=e^{\int \frac{1}{1+y^2} d y}=e^{\tan { }^2 y}
$$
$\therefore$ Solution is given by
$$
\begin{aligned}
& x \cdot e^{\tan y^{\prime} y}=\int \frac{\tan ^2 y}{1+y^2} e^{\tan y^{\prime}} d y+k \\
& \Rightarrow \quad x \cdot e^{\tan ^{\prime} y}=\int t \cdot e^t d t+k \quad\left[\text { where, } \tan ^2 y=t\right. \text { ] } \\
& \Rightarrow \quad x \cdot e^{\tan { }^{\prime} y}=t e^t-e^t+k \\
& \Rightarrow \quad x e^{\tan ^2 y}=e^{\tan ^2 y}\left(\tan ^3 y-1\right)+k \\
& \Rightarrow \quad x=\left(\tan ^2 y-1\right)+k e^{\tan ^3 y} \\
&
\end{aligned}
$$