What is this problem called and is it hard? given $g^x$ output ($g^y, xy$)
This problem is equivalent to the CDH problem:
Here is how to solve CDH given an Oracle that solves this problem:

Given $g, g^x$, we can compute $g^{x^{-1}}$ (which is equivalent to the CDH problem) by doing the following:

Call the Oracle with $g, g^x$; the Oracle gives us a pair $g^{y}, xy$

We compute $(g^{y})^{(xy)^{-1}} = g^{x^{-1}}$, hence showing one side of the equivalence


And, repeating my comment, here is how to solve this problem given a CDH Oracle:

Given $g, g^x$, we select an arbitrary value $z = xy$, and compute $(g^x)^{z^{-1}} = g^{y^{-1}}$.

We then invoke the CDH Oracle with $g, g^{y^{-1}}$ to recover the value $g^y$

We then return the values $g^y, z$, which shows the other side of the equivalence.