This equation is actually the solution to the intersection of the two surfaces $z = x^2 - y^2$, and $z = x^2 + xy - 1$.
I am to parametrise the solution curve, which is noted in the title. A first attempt is below:
\begin{align} z &= x^2 - y^2 = x^2 + xy - 1 &\\ \implies x^2 - y^2 &= x^2 + xy - 1 &\\ -y^2 &= xy - 1 &\\ xy &= y^2 - 1 \end{align}
Now, to parametrise, i.e. $x = f(t)$, $y = f(t)$: \begin{align} xy &= y^2 - 1 &\\ x &= \frac{y^2 - 1}{y} &\\ &= y - \frac{1}{y} \end{align}
Let $y = t \in \mathbb{R}\backslash0$, and hence we have: \begin{cases} x = t - \frac{1}{t} \\[2ex] y = t \end{cases}
Will that suffice? I think not; there must be something more.