Explanation

(b)
Given differential equation can be rewritten as
$\frac{y}{(1 + y^{2})}dy = \frac{dx}{x(1 + x^{2})}$
$\Rightarrow$ $\frac{1}{2}\ \int_{}^{}\frac{2y}{(1 + y^{2})}dy = \frac{1}{2}\ \int_{}^{}\frac{2x}{x^{2}(1 + x^{2})}dx$
$\Rightarrow$ $\frac{1}{2}\ \int_{}^{}\frac{2y}{(1 + y^{2})}dy = \frac{1}{2}\ \int_{}^{}\frac{dt}{t(1 + t)}$
[put $x^{2} = t$ in RHS integral]
$\Rightarrow$ $\frac{1}{2}\ \int_{}^{}\frac{2y\ dy}{1 + y^{2}} = \frac{1}{2}\ \int_{}^{}\left( \frac{1}{t} - \frac{1}{1 + t} \right)dt\$
$\Rightarrow$ $\frac{1}{2}\log{(1 + y^{2})} = \frac{1}{2}\left\lbrack \log t - \log(1 + t) \right\rbrack + \frac{1}{2}\log c$
$\Rightarrow$ $\log{(1 + y^{2})} = \log x^{2} - \log{(1 + x^{2})} + \log c$
$\Rightarrow$ $\log{\left( 1 + y^{2} \right)(1 + x^{2})} = \log{cx^{2}}$
$\Rightarrow$ $\left( 1 + y^{2} \right)\left( 1 + x^{2} \right) = cx^{2}$