This is similar to question A1 on the 2001 Putnam Exam, though the solution is a little different. You just have to play around a bit (or a lot, if you're new to this sort of thing):
Claim 1: $(x\star y)\star x=y$ for all $x,y\in X$. Proof: Let $x,y\in X$ be arbitrary. Then, $(x\star y)\star y=x$ by assumption. Now left-multiply by $(x\star y)$ to see that $(x\star y)\star((x\star y)\star y)=(x\star y)\star x$. Now apply your second property to the left hand side, with $(x\star y)$ taking the role of $y$, to see that the left hand side is actually just $y$. Thus $y=(x\star y)\star x$.
Now, since $(x\star y)\star x=y$ for all $x,y$, simply multiply on the right by $x$ to obtain $((x\star y)\star x)\star x=y\star x$
Use your first property with $(x\star y)$ playing the role of $x$ and $x$ playing the role of $y$, and you're done!