If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{d y}{d x}$ is equal to

$-\frac{y}{x}$

We have,

$x^2+y^2=t+\frac{1}{t}$  and  $x^4+y^4=t^2+\frac{1}{t^2}$

$\Rightarrow \left(x^2+y^2\right)^2=\left(t+\frac{1}{t}\right)^2$

$\Rightarrow x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$

$\Rightarrow x^4+y^4+2 x^2 y^2=x^4+y^4+2$

$\Rightarrow x^2 y^2=1 \Rightarrow 2 x y^2+2 x^2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{y}{x}$