SOLUTION —
$\begin{array}{l}\text { } \frac{x^2-y^2}{x^2+y^2}=\sec ^{-1}\left(e^a\right) \\\frac{\left(x^2+y^2\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^2-y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^2+y^2\right)^2}=0 \\\Rightarrow \quad x\left(x^2+y^2\right)-y\left(x^2+y^2\right) \frac{d y}{d x} \\=\left(x^2-y^2\right) x+y\left(x^2-y^2\right) \frac{d y}{d x} \\\Rightarrow \quad \quad \quad \frac{d y}{d x}=\frac{2 x y^2}{2 x^2 y}=\frac{y}{x}\end{array}$
So, The correct option will be (B).