Answer
$$r(t)=(\frac{1}{t}-t, t,\frac{1}{t^2}-2).$$
Work Step by Step
We have $z=x^2-y^2$ and $z=x^2+xy-1$.
Hence
$$x^2-y^2=x^2+xy-1\Longrightarrow xy=1-y^2\Longrightarrow x=\frac{1}{y}-y.$$
Also, $$z=x^2-y^2\Longrightarrow z=(\frac{1}{y}-y)^2-y^2=\frac{1}{y^2}-2.$$
Now let $y=t$; then we have $x=\frac{1}{t}-t$ and $z=\frac{1}{t^2}-2$. So we have the parametrization
$$r(t)=(\frac{1}{t}-t, t,\frac{1}{t^2}-2).$$
