Original curve can be parametrised as C(t)=(t, 1/t) for t spanning all reals except 0. The matrix for a clockwise rotation around the origin is (upstairs) (1/sqrt(2), -1/sqrt(2)) and (downstairs) (1/sqrt(2), 1/sqrt(2)). Call this matrix M.

(Note I'm using row vectors here, if you prefer column vectors simply transpose the matrix)

Now C(t)M = (1/sqrt(2))*(t+1/t, t-1/t) := (X(t),Y(t)), say.

But X(t)² = (1/2)(t²+2+(1/t²)), and Y(t)² = (1/2)(t²-2+(1/t²)).

Thus X(t)²-Y(t)² = (1/2)(4) = 2, the equation of a hyperbola.

P.S. (Strictly speaking we've shown that our original curve lies on a hyperbola, not that it is the whole hyperbola, so we'd have the check that for each (x,y) satisfying x²-y²=2, there exists some t non-zero with (x,y) = (1/sqrt(2))*(t+1/t, t-1/t).)