If $x=a\left(t-\frac{1}{t}\right), y=b\left(t+\frac{1}{t}\right)$, then $\frac{d y}{d x}=$

$\frac{b^2 x}{a^2 y}$

$x=a\left[t-\frac{1}{t}\right]$,       $y=b[t+\frac{1}{t}]$

$x^2=a^2\left[t^2+\frac{1}{t^2}-2\right]$       

$y^2=b^2\left[t^2+\frac{1}{t^2}+2\right]$

so  $\frac{y^2}{b^2}-\frac{x^2}{a^2}=4$

differentiating eq wrt (i)

$\frac{2 y}{b^2} \frac{d y}{d x}-\frac{2 x}{a^2}=0$

so  $\frac{2 y}{b^2} \frac{d y}{d x} =\frac{2 x}{a^2}$

so  $\frac{d y}{d x}=\frac{b^2 x}{a^2 y}$